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MATHEMATICS 
L'BEARY 


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_ Latest Date stamped below. 


University of Illinois Library 


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UNIVERSITY OF ILLINOIS LIBRARY 


NOV 22 


ALGEBRA FOR BEGINNERS. 


ALGEBRA 


“OR BEGINNERS 


BY ae | eke 
ay vv at 
‘abe 's! “HALL anv st Rs KNIGHT, 
’ Autuors or “ELEMENTARY ALGEBRA FOR Scnootns,”’ ‘* Hiener 


ALGEBRA,”’ ‘‘ ELEMENTARY TRIGONOMETRY,’ Ero. Ero. 


REVISED AND ADAPTED TO AMERICAN SCHOOLS 
BY 


FRANK L. SEVENOAK, A.M., M.D.,. 


PROFESSOR OF MATHEMATICS AND ASSISTANT PRINCIPAL IN 
THE STEVENS SCHOOL, ACADEMIC DEPARTMENT OF 
THE STEVENS INSTITUTE OF TECILNOLOGY 


New Work 
MACMILLAN & CO. 
AND LONDON 


1895 


All rights reserved 


CopyricuT, 1895, 


By MACMILLAN & CO. 


Norwood Wpress : 
J. S. Cushing & Co. — Berwick & Smith. 
Norwood, Mass., U.S.A. 


_ f") 
= fae fl 
ated Ba ad Bs SS 


H\4a JATNEMATICS LIBRARY 


PREFACE. 


Tur rearrangement of the Elementary Algebra of 
Messrs. Hall and Knight was undertaken in the hope 
of being able to give to our advanced secondary schools 
a work that would fully meet their requirements in this 
important study. Many changes were made and ad- 
ditional subject-matter introduced. The Algebra, for 

. Beginners, however, so fully meets the needs of the 
tselass of students for which it was written, that we have 
made only such changes as seemed to bring out more 
clearly important points, and better adapt it to American 
schools. 

With reference to the arrangement of topics, we quote 
from Messrs. Hall and Knight’s preface to a former 
edition : 

Y “Our order has been determined mainly by two con- 
*’ siderations: first, a desire to introduce as early as pos- 

‘sible the practical side of the subject, and some of its 

most interesting applications, such as easy equations and 
~“eproblems; and secondly, the strong opinion that all 
%sreference to compound expressions and their resolution 

nto factors should be postponed until the usual opera- 
tions of Algebra have been exemplified in the case of 
simple expressions. By this course the beginner soon 


va. 351058 


thang 


& Mid = 


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oy & 


- 


v1 PREFACE. 


becomes acquainted with the ordinary algebraical proc- 
esses without encountering too many of their difficulties ; 
and he is learning at the same time something of the 
more attractive parts of the subject. 

“ As regards the early introduction of simple equations 
and problems, the experience of teachers favors the 
opinion that it is not wise to take a young learner 
through all the somewhat mechanical rules of Factors, 
Highest Common Factor, Lowest Common Multiple, 
Involution, Evolution, and the various types of Frac- 
tions, before making some effort to arouse his interest 
and intelligence through the medium of easy equations 
and problems. Moreover, this view has been amply sup- 
ported by all the best text-books on Elementary Algebra 
which have been recently published.” 

The work will be found to meet the wants of all who 
do not require a knowledge of Algebra beyond Quadratic 
Equations —that portion of the subject usually covered 
in the examination for admission to the classical course 
of American Colleges. 

FRANK L. SEVENOAK. 

JUNE, 1895. 


= : arn 


' 
OHAP. PAGE 
I. DEFINITIONS. SUBSTITUTIONS . : : : : ; o 
pis NEGATIVE QUANTITIES. ADDITION OF Likk TERMS : i 


yi II. Simphe Brackets.» ADDITION. .. + + DS ee ce LL 
" IV. SUBTRACTION . ; “. . ‘ . " ° 2 16 
MISCELLANEOUS EXAMPLES I. . . . . ° ; 19 

V. MULTIPLICATION ; ; ; - P c : : 21 

wae, Vi, DIVISION, . : - ‘ a : : ‘ ° ‘ 31 


« ¥ 


fp __NI1...REMOVAL AND INSERTION OF BRACKETS . F $ A 38 


Seo MISCELLANEOUS EXAMPLES II, . : : : r 42 
VIIl. BReEvIsION oF ELEMENTARY RULES . : : ‘ 3 44 
IX. SiMPLE EQUATIONS . ; Se ‘ : : - 52 
X. SyYMBOLICAL EXPRESSION . ‘ . : : : ‘ 59 


XI. Propiems LEADING TO SIMPLE EQUATIONS . F A 64 


>. 


XII. Hicuest Common Factor, Lowest Common MULTIPLE 


or Simple EXPRESSIONS. FRACTIONS INVOLVING 


SIMPLE EXPRESSIONS ONLY. : : ; . : (fs 
XIII. SimuULTANEOUS EQUATIONS : ‘ : P H ; 17 
XIV. PROBLEMS LEADING TO SIMULTANEOUS EQuaTIons : 85 
XV. INVOLUTION ; : ‘ ; : , ‘ ; 89 
XVI. EvoOLUTION ; ; : : 5 : * : : 93 


ry 


XVII. RESOLUTION INTO FACTORS. CONVERSE USE OF FACTORS 101 


MISCELLANEOUS EXAMPLeEs III. ‘ ; ; . Peg BE 


XXV, 


XXXVI. 


CONTENTS. 


PAGE 
HIGHEST COMMON FACTOR OF COMPOUND EXPRESSIONS 116 


MULTIPLICATION AND DIVISION OF FRACTIONS ; ; 122 


LOWEST COMMON MULTIPLE OF CoMPOUND EXPRESSIONS 126 


. ADDITION AND SUBTRACTION OF FRACTIONS . : fF Wea) 
MISCELLANEOUS FRACTIONS . .. : ; : . 139 
HARDER EQUATIONS . : 5 ; ; : ; . 146 
HARDER PROBLEMS . ; ; ° . ° : . 153 
Misce.taNrous Exampims IV. ee oe 
QUADRATIC EQUATIONS . : ‘ ‘ ° ne LG2 
PROBLEMS LEADING TO QUADRATIC EQUATIONS. Lis 


MISCELLANEOUS EXAMPLES V, : ‘ 4 : alee 


ALGEBRA. 


CHAPTER L 


DEFINITIONS. SUBSTITUTIONS. 


1, ALGEBRA treats of quantities as in Arithmetic, but with 
greater generality ; for while the quantities used in arithmetical 
processes are denoted by figures which have one single definite 
value, aleebraical quantities are denoted by symbols which may 
have any value we choose to assign to them. 


The symbols employed are letters, usually those of our own. 
alphabet ; and, though there is no restriction as to the numerical 
values a symbol may represent, it is understood that in the same 
piece of work it keeps the same value throughout. Thus, when 
we say “let a=1,” we do not mean that a must have the value 
1 always, but only in the particular example we are considering. 
Moreover, we may operate with symbols without assigning to 
them any particular numerical value at all; indeed it is with 
such operations that Algebra is chiefly concerned. 


We begin with the definitions of Algebra, premising that the 


symbols +, —, x, +, will have the same meanings as in 
Arithmetic. 





2. An algebraical expression is a collection of symbols; 

it may consist of one or more terms, which are separated from 

each other by the signs + and —. Thus 7a+5b—38c-—4#+2y is 
an expression consisting of five terms. 


Note. When no sign precedes a term the sign + is understood. 


3, Expressions are either simple or compound. A. simple 
expression consists of one term, as 5a. A compound expression 
consists of two or more terms. Compound expressions may be 

€ H.A. A 


2 ALGEBRA. [CHAP. 


further distinguished. Thus an expression of two terms, as 
3a — 26, is called a binomial expression ; one of three terms, as 
2a—3b+c¢, a trinomial; one of more than three terms a multi- 
nomial. 


4, When two or more quantities are multiplied together the 
result is called the product. One important difference between 
the notation of Arithmetic and Algebra should be here remarked. 

#3;In Arithmetic the product of 2 and 3 is written 2 x 3, whereas 
in Algebra the product of a and 6 may be written in any of 
the forms axb, a.6, or ab. The form ab is the most usual. 
Thus, if a=2, b=3, the product ab=axb=2x3=6; but in 
Arithmetic 23 means “twenty-three,” or 2x 10+3. 


5, Each of the quantities multiplied together to form a pro- 
duct is called a factor of the product. Thus 5, a, b are the 
factors of the product 5ab. 


6, When one of the factors of an expression is a numerical 
quantity, it is called the coefficient of the remaining factors. 
Thus in the expression 5a), 5 is the coefficient. But the word 
coefficient is also used in a wider sense, and it is sometimes 
convenient to consider any factor, or factors, of a product as 
the coefficient of the remaining factors. Thus in the product 
Gubc, 6a may be appropriately called the coefficient of be. A 
coefficient which is not merely numerical is sometimes called a 
literal coefficient. 

Note. When the coefficient is unity it is usually omitted. Thus 
we do not write la, but simply a. 

7, If a quantity be multipled by itself any number of times, 
the product is called a power of that quantity, and is expressed 
by writing the number of factors to the right of the quantity 
and above it. Thus 

axa is called the second power of a, and is written a? ; 
LS URIS ere seas setter, POWE? OL Cau. wherne metic: fe a’ ; 
and so on. 

The number which expresses the power of any quantity is 
called its index or exponent, Thus 2, 5, 7 are respectively 
the indices of a”, a®, a’. 

Note. a? is usually read ‘“‘a squared”; a® is read ‘fa cubed”; 
a* is read ‘‘a@ to the fourth” ; and so on. 


When the index is unity it is omitted, and we do not write 
a', but simply a. Thus a, la, a’, 1a4 all have the same meaning. 


I] DEFINITIONS. SUBSTITUTIONS. 2 


8. The beginner must be careful to distinguish between 
coefficient and index. 


Lzxample 1, Whatis the difference in meaning between 3a and a?? 
By 3a we mean the product of the quantities 3 and a, 
By a? we mean the third power of a; that is, the product of the 
quantities a, a, a. 
Thus, if a = 4, 
34:2 3X GS KAS 


a=axaxa=4x4x4= G4. 


Example 2. If b =5, distinguish between 4b? and 2b4+, 
Here 4b =A b= 4x xX B= 1003 
whereas OU Ue D0 = DOR Oo MOK D = 1250. 


Hxample 3. If «x =1, find the value of 52%. 
Here De = Mo woe x om Le heel =D; 


Note. The beginner should observe that every power cf 1 is 1. 


9, In arithmetical multiplication the order in which the 
factors of a product are written is immaterial. For instance 
3x4 means 4 sets of 3 units, and 4x3 means 3 sets of 4 units; 
in each case we have 12 units inall. Thus 


3x4=4x3, 
In a similar way, 
Ox4 xX Daat x ox Dat SD Mo 
and it is easy to see that the same principle holds for the 
product of any number of arithmetical quantities. 


In like manner in Algebra ab and ba each denote the product 
of the two quantities represented by the letters a and b, and 
have therefore the same value, Again, the expressions abc, 
ach, bac, bca, cab, cba have the same value, each denoting the 
product of the three quantities a, 6, c. It is immaterial in 
what order the factors of a product are written; it is usual, 
however, to arrange them in alphabetical order. 


Fractional coefficients which are greater than unity are 
usually kept in the form of improper fractions. 


Example 4. If a= 6, # = 7, z= 5, find the value of ‘as 


Here Paws = 3x 6x7 x d= 273, 


4 ALGEBRA. [crrap. 


EXAMPLES I. a. 

Ifa=5, (bes oe lee 3, y= 12"2-2 and thevalue of 

Le ed. Ne ibe Oreos Ate. Ny eae 

Geese Tye ab Se ce Ne a ARGs Oe 
Lee Lp, Melons 18, 62°. TAS be. 15, Fe: 

1b re 6, p =4,q=7, r=5, 2 =1, find the value of - 
16 Sap. Lye opG: IG SCE, 1D eOTe: 90, S8aqa. 
D1. por. OD SA0T eS DG 1 OTe: Pe le TANS | bas 
DOs op. WA Hei 08> Dapon 29 556xro ith, een 


lfih=5, k=3, 2 =4, y =], find the value of 


1 1 The 1 pclos'f 
\ps. Ve, La, 1 ike. 1 ys, 
Bl. G6, 82, abe 88, fy 8A, hk, 85, Sy 
Chee co yeeoe op ke yee og tis «Ag Shee 
* 8 * OF “7125 sa ys 


10, When several different quantities are multiplied together ~ 
a notation similar to that of Art. 7 is adopted. Thus aabbbbeddd 
is written a?b’cd>. And conversely 7a*cd? has the same meaning 
as 7xaxaxaxcxdxd. 

Example 1. If c=3, d=, find the value of 16c*a?. 

Here. 16ctP S16 x 34x? = (16 x5") 3 | 2000 81 62000: 

Note. The beginner should observe that by a suitable combination 
of the factors some labour has been avoided. 

Example2, If p=4, q=9, r=6, s=5, find the value of oa 


Slps* 
ager 32q7r3_382x9x6?_ 32x 9x6x6x6 _3 
Slp? = 81x4® S8lx4x4x4x4x4 4 





11, If one factor of a product is equal to 0, the product must 
be equal to 0, whatever values the other factors may have. A 
factor O is usually called a zero factor. 

For instance, if v=0 then ab’ry? contains a zero factor. 
Therefore ab?xy? =0 when «=0, whatever be the values of a, b, 7. 

Again, if c=0, then c?=0; therefore ab’c}=0, whatever values 
aand b may have. 


Note. Every power of 0 is 0. 


ot 


Tet DEFINITIONS. SUBSTITUTIONS. 


EXAMPLES I. b. 
feos pec =) 1 0, 2=7, find the value of 


1, 3bp. O- Bax. 3, op. 4, Gaqz. 5, bpz. 
6, 3079. pease Segre. 9, gz’. 10, 50? pa. 
Vive tee oeop 7. - lomeeas 1S ares 15, Saiq’. 


ieee = 2.90 = 0,-p = 3,'¢ = 45.97=.5, find the value of 




















3k? 513 m 3m? 16p? 
16, p? he qr’ 18, Bye 19, 4/ $ 20. Iq" . 
5m? 60 8r3 Imq 8lqtr? 
als es a OO: 3q? 23. 25lq? 94, 4p? 25, 400p> 
ma JN q® kr* 5m” ee 
26, aT. G08, a9, S80, 


12, We now proceed to find the numerical value of expres- 
sions which contain more than one term. In these each term 
can be dealt with singly by the rules already given, and by 
combining the terms the numerical value of the whole expres- 
sion is obtained. 


18. We have already, in Art. 8, drawn attention to the 
importance of carefully distinguishing between coefficient and 
index ; confusion between these is such a fruitful source of 
error with beginners that it may not be unnecessary once more 
to dwell on the distinction. 


Hxample. When c=5, find the value of ct ~ 4c + 2c? — 8c?. 
Here (oe OGD KO on 
dee 4 = 2) « 
Qe? 2 5) 2 6 DX = 250 + 
BC = 8X07 KD 
Hence the value of the expression 
= 625 — 20 + 250-75 = 780. 


14, The beginner must also note the distinction in meaning 
between the swm and the product of two or more algebraical 
quantities. For instance, ab is the product of the two quan- 
tities a and 0b, and its value is obtained by multiplying them 
together. But a+b is the sum of the two quantities a and 8, 
and its value is obtained by adding them together. 


6 ALGEBRA. [CHAP. 1. 


Thus if a=11, 6=12, 
the sum of a and b is 11412, that is, 23 ; 
the product of a and b is 11x12, that is, 132. 


15, By Art. 11 any term which contains a zero factor is 
itself zero, and may be called a zero term. 


Example. Ifia=2,b=0, x=5, y =3, find the value of 
5a’ — ab? + 2au°y + 8bay. 
The expression = (5 x 2?) -0+ (2x 5*x 3)+0 
= 40+ 150 = 190. 
Note. The two zero terms do not affect the result. 


16, In working examples the student should pay attention 
to the following hints. 


1. Too much importance cannot be attached to neatness of 
style and arrangement. The beginner should remember that 
neatness is in itself conducive to accuracy. 


2. The sign = should never be used except to connect 
-,e . . . 
quantities which are equal. Beginners should be particularly 
careful not to employ the sign of equality in any vague and 

mexact sense. 
3. Unless the expressions are very short the signs of equality 
in the steps of the work should be placed one under the other. 


4. It should be clearly brought out how each step follows 
from the one before it; for this purpose it will sometimes be 
advisable to add short verbal explanations ; the importance of 
this will be seen later. 


EXAMPLES I. c. 
lig=4,b=21,0=3;f—o0.9 = 1, d= 0, find themalue of 


1, 3/+5h- 7b. 2, Tco-9h+2a. 38, 49-5c-9b. 
4, 3g-4h+7c. 5, 3f-29-6. 6, 9b-3c+4h. 
7, 3a-9b+e. 8, 2f-39+5a. Q, 3c-—4a+7b. 
10, 3/+5h-2c-4b+a. ll, 6h-7b-5a-7f+99. 
12, 7c+5b —4a+Sh+3g. 13, 9b+a-394+4f+7h. 


14, fgtgh-ab. 15, gb-3he+fb. 16, fht+hb-3he. 
Ll]. f7*= 3a" 4.2¢7, 18, 0?-2h?+3a?, 19, 3b?-2b?+ 4h? - 2h. 


CHAPTER II. 


NEGATIVE QUANTITIES. ADDITION OF LIKE TERMS. 


17, Iv his arithmetical work the student has been accus- 
tomed to deal with numerical quantities connected by the signs 
+ and — ; and in finding the value of an expression such as 
17+72-—344+6-41 he understands that the quantities to which 
the sign + is prefixed are additive, and those to which the sign 
— is prefixed are subtractive, while the first quantity, 13, to 
which no sign is prefixed, is counted among the additive terms. 
The same notions prevail in Algebra ; thus in using the expres- 
sion 74+3b—4c—2d we understand the symbols 7a and 36 to 
represent additive quantities, while 4c and 2d are subtractive. 


18, In Arithmetic the sum of the additive terms is always 
greater than the sum of the subtractive terms; if the reverse 
were the case the result would have no arithmetical meaning. 
In Algebra, however, not only may the sum of the subtractive 
terms exceed that of the additive, but a subtractive term may 
stand alone, and yet have a meaning quite intelligible. 


Hence all algebraical quantities may be divided into positive 
quantities and negative quantities, according as they are 
expressed with the sign + or the sign — ; and this is quite 
irrespective of any actual process of addition and subtraction. 


This idea may be made clearer by one or two simple illus- 
trations. 


G) Suppose a man were to gain $100 and then lose $70, his 
total gain would be $30. But if he first gains $70 and then 
loses $100 the result of his trading is a loss of $30. 


The corresponding algebraical statements would be 
$100—$70 = +30, 
$70 —$100 = — $30, 


8 ALGEBRA. [CHAP. 


and the negative quantity in the second case is interpreted as a 
debt, that is, a sum of money opposite in character to the positive 
quantity, or gain, in the first case; in fact it may be said to 
possess a subtractive quality which would produce its effect 
on other transactions, or perhaps wholly counterbalance a sum 
gained, 


(1) Suppose a man starting from a given point were to walk 
along a straight road 100 yards forwards and then 70 yards 
backwards, his distance from the starting-point would be 30 
yards. But if he first walks 70 yards forwards and then 100 
yards backwards his distance from the starting-point would be 
30 yards, but on the opposite side of it. As before we have 


100 yards— 70 yards= +30 yards, 
70 yards —100 yards= — 30 yards. 


In each of these cases the man’s absolute distance from the 
starting point is the same; but by taking the positive and 
negative e signs into account, we see that —30 is a distance from 
the starting point equal in magnitude but opposite in direction 
to the distance represented by +30. Thus the negative sign 


nay here be taken as indicatir : l of direction. 
may here be taken as indicating a reversal of direct 


(iii) The freezing point of the Centigrade De ae is 
marked zero, and a temperature of 15° C. means 15° above the 
freezing point, while a temperature 15° below the freezing point 
is indicated by — 15° C. 


19. Many other illustrations might be chosen; but it will! 
be sufficient here to remind the student that a subtractive 
quantity is always opposite in character to an additive quantity 
of equal absolute value. In other words subtraction ts the reverse 
of addition. 


20, Derinition. When terms do not differ, or when they 
differ only in their numerical coefficients, they are called like, 
otherwise they are called unlike. Thus 3a, 7a; 5a°b, 2u*b ; 

3a°b?, —4a°b? are pairs of like terms; and 4a, 3b ; “Te? 9a°b are 


pairs of unlike terms. 


Addition of Like Terms. 


RuleI. The sum of a number of like terms is a like term. 
Rule II, Jf all the terms are positive, add the coefficients. 


u.] NEGATIVE QUANTITIES. ADDITION OF LIKE TERMS. 9 


Example. Find the value of 8a + 5a. 


Here we have to increase § like things by 5 like things of the 
same kind, and the aggregate is 13 of such things ; 


for instance, 8 lbs. + 5 lbs. = 13 lbs. 
Hence also, 8a+5a = 13a. 
Similarly, 8a+5a+a+2a+6a = 22a, 


Rule III. Jf all the terms are negative, add the coefficients 
numerically and prefix the minus sign to the sum. 


Example. To find the sum of -3a, -—5a, —7x, —2. 


Here the word sum indicates the aggregate of 4 subtractive 
quantities of like character. In other words, we have to take away 
successively 3, 5, 7, 1 like things, and the result is the same as 
taking away 3+5+7+1 such things in the aggregate. 


Thus the sum of —3x, —5x, -—7x, -—x is — 162. 


Rule IV. Jf the terms are not all of the same sign, add to- 
gether separately the coefficients of all the positive terms and the 
coefficients of all the negative terms; the difference of these two 
results, preceded by the sign of the greater, will give the coefficient 
of the sum required. 


Example 1. The sum of 17% and —8z is 9x, for the difference of 
17 and 8 is 9, and the greater is positive. 


Example 2. To find the sum of 8a, -9a, —a, 3a, 4a, —lla, a. 


The sum of the coefficients of the positive terms is 16. 
PS ee rege cts ef cis yer es Puce vckies negative.......... 21. 


The difference of these is 5, and the sign of the greater is nega- 
tive; hence the required sum is — 5a. 


We need not however adhere strictly to this rule, for the 
terms may be added or subtracted in the order we find most 
convenient. 


This process is called collecting terms. 


‘21, When quantities are connected by the signs + and -, 
the resulting expression is called their algebraical sum. 

Thus lla—27a+13a=-— 3a states that the algebraical sum 
of lla, —27a, 13a is equal te — 8a, 


22. The sum of two quantities numerically equal but with 
opposite signs is zero. ‘Thus the sum of 5a and — du is 0. 


10 


ALGEBRA. 


(CHAP. II. 


EXAMPLES II. 


Find the sum of 


COTS Si Con 


11, 
13, 
15, 
A: 
19, 


2a, 3a, 6a, a, 4a. 

Gb, 11b, 8b, 9b, 5b. 

2p, p, 4p, Tp, 6p, 12p. 
—2x, —62, —10a, - S82. 
Uy 49) ou, OY ae 
—2ly, —5y, -8y, - 18y. 
-—4s, 3s, s, 2s, -2s, —-s. 
32, 102, — 7x, 12x, Oa. 
2ay, —4xy, — 32y, xy, Txy. 


abe, - 8abe, 2abe, —5abce. 


Find the value of 


21, 
23, 
25. 
27. 
29. 


— 9a? + lla? +3a2-—4a%. ¢ 


~ Lia? + 3a*>— 8a? — Fa? + 2a? 


a?b? — 7a7b? + Sa2b? + 9a2b?. 
2p°q? - 81 p3q? + 17 pq’. 
9Jabed — llabcd — 4labed. 


AY 2, oe, OL, O2. 

6c, 7c, 3c, 16c, 18c, 101c. 

d, 9d, 3d, 7d, 4d, 6d, 10d. 
— 3b, -—13b, —19b, — 5b. 
—l%c, —34¢, —9c, — 6c. 
—4m, —138m, -17m, — 59m. 
LLY Oy yoy, TY: 
Sab, —6ab, 5ab, —4ab. 
Spq, —8pq, 8pq, —4p¢. 

— xyz, -Qxyz, Txy2z, — xyz. 


3b? — 2b? +76? — 96°. 

92° — 32° — Oa? — 923. 

ae — lla®x + 3a2x — Qu?a. 
Tmin —15m4n + 3m4n. 


l3pqu — d5xpq —19qpz. 


CHAPTER IM: 


SIMPLE BRACKETS. ADDITION. 


93, WHEN a number of arithmetical quantities are connected 
together by the signs + and —, the value of the result is the 
same in whatever order the terms are taken. This also holds 

in the case of algebraical quantities. 


Thus a—b+c is equivalent to a+c-—), for in the first of the 

two expressions 6 is taken from a, and ¢ added to the result ; in 

_ the second ¢ is added to a, and 6 taken from the result. Similar 

reasoning applies to all algebraical expressions. Hence we 
may write the terms of an expression in any order we please. 


Thus it appears that the expression a—6 may be written in 
the equivalent form —b+a. 


To illustrate this we may suppose, as in Art. 18, that a rep- 
resents a gain of a dollars, and —b a loss of 0 dollars: it is 
clearly immaterial whether the gain precedes the loss, or the 
loss precedes the gain. 









_ 24, Brackets ( ) are used to indicate that the terms enclosed 
within them are to be considered as one quantity. The full use 
_ of brackets will be considered in Chap. vir.; here we shall deal 
only with the simpler cases. 
8+(13+5) means that 13 and 5 are to be added and their 
sum added to 8. It is clear that 13 and 5 may be added 
separately or together without altering the result. 
Thus 8+(18+5)=84+ 13+5=26. 
Similarly a+(b+c) means that the sum of b and ¢ is to be 
added to a. | 
Thus a+(b+c)=a+b+e. 
8+(13—5) means that to 8 we are to add the excess of 13 over 
5; now if we add 13 to 8 we have added 5 too much, and must 
therefore take 5 from the result. 
Thus 8+(13—5)=8+4+13 -5=16. 
. Similarly a+(b-—c) means that to a we are to add 6, diminished 
vy c. : 
Thus at+(b-—c)=at+b-e., 


12 ALGEBRA. [oHap. 


In like manner, 
a+b—c+(d—e-f)=a+b—-c+d-—e-f. 
By considering these results we are led to the following rule : 


Rule, When an expression within brackets is preceded by the 
sign +, the brackets can be removed without making any change in 
the expression. 


25, The expression a-(b+c) means that from a we are to 
take the sum of band c. The result will be the same whether 
b and ¢ are subtracted separately or in one sum. Thus 

a—(b+c)=a-—b-c. 

Again, a—(b—c) means that from a we are to subtract the 
excess of b over c. If from a we take b we get a—b; but by so 
doing we shall have taken away ¢ too much, and must therefore 
add etoa—b. Thus 

a—(b-—c)=a—b+e. 
In like manner, 


a—b—(c—d-—e)=a—b—c+d+te. 
Accordingly the following rule may be enunciated : 


Rule. When an expression within brackets, is preceded by the 
sign —, the brackets may be removed if the sign of every term within 
the brackets be changed. 


Addition of Unlike Terms. 


26, When two or more like terms are to be added together we 
have seen that they may be collected and the result expressed 
as a single like term. If, however, the terms are unlike they 
cannot be collected; thus in finding the sum of two unlike 
auantities a and 6, all that can be done is to connect them by 
the sign of addition and leave the result in the form a+b. 


27, We have now to consider the meaning of an expression 
like a+(—6). Here we have to find the result of taking a 
negative quantity —b together with a positive quantity a. 


Now —b implies a decrease, and to add it to @ is the ¢ /.1e in 
effect as to subtract 6; thus 


a+(-—b)=a—6b; 


that is, the algebraical sum of a and —b is expressed by a—b. 


98, It will be observed that in Algebra the word swm is used 
in a wider sense than in Arithmetic. Thus, in the language of 
Arithmetic, a—6 signifies that 6b is to be subtracted from a, 


III. } ADDITION. 13 


and bears that meaning only ; but in Algebra it is also taken 
to mean the sum of the two quantities a and —b without any 
regard to the relative magnitudes of a and 0. 


Example 1. Find the sum of 38a —-5b+2c, 2a+3b-d, -—4a+ 20. 


The sum = (3a — 5b + 2c) + (2a +386 - d) +(—4a+ 2b) 
= 38a -5b4+2c+2a4+3b-—d-4a+2b 
= 8a+ 2a -4a—-5b4+3b+2b+2c-d 
=a+2c-d, 
by collecting like terms. 


The addition is however more conveniently effected by the 
following rule : 


Rule. Arrange the expressions in lines so that the like terms 
may be in the same vertical columns: then add each column 
beginning with that on the left. 





3a — 5b + 2c The algebraical sum of the terms in the 
2a +3b ayy first column is a, that of the terms in the 
ep ee second column is Zero, The single terms 
in the third and fourth columns are 

a +2c-d brought down without change. 


Hxample 2. Add together -5ab+6bc-TJac; 8ab+3ac-2ad; 
~2ab+4act+5ad; be-3ub+4ad. 
— 5ab + 6bce — Tac 
Sab + 3ac —-2ad Here we first rearrange the ex- 
“Ge LIES pressions so that like terms are in 
r the same vertical columns, and then 
~ 8ab+ be +4ad add up each column separately. 
— 2ab + Tbe +7ad 





EXAMPLES III. a. 


Find the sum of 


3a+2b-5c¢; —4a+b—Te; 4a-3b+6c. 
32+2y+625; w2-3y-32; Qa+y—3z. 
4p+3q+dr; -2p+8q—-8r; p-qt+r. 
7a—5b+8c3; lla+2b-c; 16a+5b-2c. 
8l-2m+5n; -614+7m+4n; -1—4m-8n. 
5a-—7b+3c—4d; 6b-5c+3d; b4+2c-d. 
2a+4b—5x; 2b-5a; -38a+2y; -G6b+Sa+y. 
Wa-Sy-Tz; 4e+y; 62; b5a-38y+2z, 


°. e « 


° 


ONaoorRrwhr 


. 


14 ALGEBRA. [CHAP, 


9, a—-2b+7c¢+3; 2b-8c4+5; 38c+2a; a-8-Te. 

10, 5-x-y; 7+2xe3; 3y-22; -4+a”-2y. 

1], 25a-15b+c; 4c-10b+18a; a-—c+20b. 

12, 2a-3b-2c+2x; 5x24+3b-7c; 9c-6x—2a. 

18, 3a-—5c+2b-2d; b+2d-a; 5c+38f+3e —2a - 3d. 
16a Pe ls POU EK Ps BOO ie eon 

15, l7ab-13kl-Say; Tay; 12kl—5ab; 3xy—4kl-ab. 
16, 2ax-38by-—2cz; 2y-ax+7cz3; axw-—4cz+7Tby; cz—Gby. 
17, Saxv+cz-4by; Thy-S8ax-cz3; -3hby+9ax. 

18, 3+5cd; 2fg-3st; 1-5ced3; -442st—fy. 

19, 5cea+38fy-2+2s; -2fy+6-9s; -3s-4+4+2cx—-fy. 
90, -—38ab+7cd-—5qr; 2ry+8qr—cd; 2cd-3qr+ab—2ry. 


29, Different powers of the same letter are unlike terms ; 
thus the result of adding together 2x”? and 3x7 cannot be ex- 
pressed by a single term, but must be left in the form 2x? + 327. 

Similarly the algebraical sum of 5a7b?, —3ab°, and —O* is 
5a°b? — 3ab—b*, This expression is in its simplest form and 
cannot be abridged. 


Example. Find the sum of 6x? — 52, 2x7, 5a, —2a°, - 32, 2. 
The sum = 62° — 54 +22? + 5a — 223 — 3x2+2 

= 623 — 2Qu3 + Qu? — 8a? -5a+5x+2 

= 47? — 7? +9, 
This result is in descending powers of 2. 


30, In adding together several algebraical expressions con- 
taining terms with different powers of the same letter, it will be 
found convenient to arrange all expressions in descending or 
ascending powers of that letter. This will be made clear by 
the following example. 


Example 1. Add together 32° +7 + 6” - 52°; 27-8 - 9a; 
Ada —2a° 4327; 32a°-9e—2?; w-a?-774+4. 


303 — 522+ 6a+7 | In writing down the first expression 
Tr ee we put in the first term the highest 

ss power of xz, in the second term the 

~ 20° + 3a? + daz next highest power, and so on till the 
30° — x2? -—9x last term in which x does not appear. 
ia dg ites peed The other expressions are arranged in 





the same way, so that in each column 


32° — 20° -Ta+3 we have like powers of the same letter. 





III. | 


ADDITION, 15 


Example 2. Add together 


8ab? - 2b? +a; 5a*b - ab? - 3a? 


— 2b? + 3ab? ea? 


5b? 





3b? 


— ab?+ 5a*b -3a3 
+ 8a? 

ab?+ 9a?b —-2a3 

+ 3ab? + 14a7b + 4a? 





; 8a2+5b?; 9a2b - 2a? + ab? 


Here each expression contains 
powers of two letters, and is 
arranged according to descend- 
ing powers of b, and ascending 
powers of a. 


EXAMPLES III. b. 


Find the sum of the following expressions : 


SON OORONH 


v7 + 3xy-38y°?; -—3a?+ay+2y?; Q2x?-3xyt+y". 


2427-2443; —247+5x4+4; 
5a —a2+a-—-1; 2x? -Qr+5; 


xz” — Qa -—6. 
~— §2° + 5a — 4. 


a®—a*b+5ab?+b?; -—a®—10ab?+6°; 2a°b+5ab?- b. 
3a? —~9x?—-lle+7; 2u?-5x?4+2; 5a2+152°-TFas 8-9. 


x°—52°+8a;3; Ta°+4a°+5e; 


822-92; 2a°- Tx? — 4a. 


4m? +2m2-5m+7; 3m?+6m?-2; -—5m?+3m; 2m-6. 
ax®—4be?+cx; 8ba?-2cxr-d; bx*+2d; Qax?+d. 


py — 99. It; = 2py" + 3qy — 


5y® + 20y2+8y-1; -2y+5- 


Gr; Tay-4r; 3py*. 
7Ty?; —8y?-4+2y8-y. 


2-—a+S8a?-a®; 2Qa?-3a2+2a-2; -—38a+4+7a3—5a? 


1+2y - 3y?-5y?; | -1+4+2y?- 
a7a3 —3a®x2-++-0; 52+ 7a®x? ; 


xv — 4darty — Bary? 3 Baty + 2a3y3 


y; 5y>+3y?+4. 
4aPu? — a?x? — 5x. 
-6xy?; 3°42 + 6ayt - y?. 


a®—4a*b+6abe; a?b-l0abe+c?; 63+3a7b+abce. 

ap’ -6bp?+Tep; 5-6cep+5bp?; 3-2ap®; 2p--7. 

e7 —20° +11c8s;  —2c7 —8c8+5c5; 4c8-10c>; 4c? -c®. 
4h3—-7+3h4-2Qh; Th-3h?+2-ht; Qh4+2Qh?-5. 

30° +4+2y?-5et+2; Tai-5y?+7x-5; 9a3+11-84%+4y?; 


6x — y? - 182° — 7. 
7+ Qry + 3y"; 327+ 2yz+y*; 
ey + 22 + yx — 62? - 4y? — 22, 


e274 32242923 22-3xy - 38yz; 


CHAPTER IV. 


SUBTRACTION. 


31, THE simplest cases of Subtraction have already come 
under the head of addition of dike terms, of which some are 
negative. [Art. 20.] 


Thus 5a-38a= 2a, 
3a —Ta= —4a, 
—da46a= — Qa. 
Since subtraction is the reverse of addition, 
+b—b=0; 
. a=at+b—b. 


Now subtract —b from the left-hand side and erase —b on 
the right ; we thus get 
a—(-—b)=a+b. 
This also follows directly from the rule for removing brackets. 
[Art. 25.] 


Thus 3a —( --5a)=38a+5a 
= 8a, 
and —3a—(-—5a)= —3a+5a 
= 2a. 


Subtraction of Unlike Terms. 


32. We may proceed as in the following example. 


Kxample. Subtract 8a—2b—c from 4a—3b+5e. 
The difference 
The expression to be subtracted is 
=4a—3b+5c—(8a—2b—c) | first enclosed in brackets with a minus 
=4a—38b)+5c—8a+2b+¢ sign prefixed, then on removal of the 
=4a—8a—3b42b+5c+¢ brackets the like terms are combined 
=a—b+6e. by the rules already exclaimed in 
Art, 20, 


CHAP. Iv.] ~ SUBTRACTION. 17 


It is, however, more convenient to arrange the work as follows, 
the signs of all the terms in the lower line being changed. 





4a, — 3b + 5e The like terms are written in 
—3a+2b+e the same vertical column, and each 
by addition, a— b+6c column is treated separately. 


Rule. Change the sign of every term in the expression to be 
subtracted, and add to the other expression. 


Note. It is not necessary that in the expression to be subtracted 
the signs should be actually changed; the operation of changing 
signs ought to be performed mentally. 


Example 1. From 5a?+ay take 2a? + 8axy - Ty. 


5at+ xy In the first column we combine mentally 52? 
and — 22°, the algebraic sum of which is 3x7, In 
= the last column the sign of the term — 77 has to 
3a" — Tay +7y? be changed before it is put down in the result. 


2a? + Say — Ty? 


Hxample 2. Subtract 3x?-2x from 1-2. 


Terms containing different powers of the same letter being wnlike 
must stand in different columns. 


_ 73 co) In the first and last columns, as there is 
nothing to be subtracted, the terms are put 
— down without change of sign. In the second 
~ 23-327 +2e+1 and third columns each sign has to be changed. 


3x? — 2x 





The re-arrangement of terms in the first line is not necessary, but 
it is convenient, because it gives the result of subtraction in descend- 
ing powers of x. 


EXAMPLES IV. 
Subtract 


a+2b—c from 2a+3b+c. 
2a—b+c from 3a—5b--c. 
3a--y-—% from 2 —4y+4 32. 
x+8y+8z from 10x —7Ty - 6z. 
—m—3n+p from -—2m-+n- 3p. 
3p -2q+r from 4p—7q+3r. 
a—7b-3c from -—4a4+35)+4+8c. 
-a-b-9c from -a+b-9e. 
H.A. B 


CONDO FPwONH 


18 ALGEBRA. [CHAP. 


Subtract 


9, 38x-5y-7z from 2x+3y —4z. 
10, -—4¢-2y+1lz from —~a#2+2y- 132. 
ll, -2x-5y from 2+ 3y- 2z. 
12, 3x-y-8z from 2+2y. 
138, m-2n-p from m+2n. 
14, 2p-3q-r from 2q-4r. 
15, ab-2cd-ac from —ab—3cd+2ac. 
16, 3ab+6cd -3ac—5ld from 3ab+5cd —4ac -- 6bd. 
17. -xyt+yz-—2x from 2xy+2. 
18, -2pq-3qr+4rs from qr -4rs. 
19, -mn+llnp-—8pm from — 1lnp. 
20, x°y -2Qxy?+ 3xyz from 2x?y + 38xy? - xyz. 


From 


91, x?-38x?+2 take —2°+322-—-2. 

99, -2x?—x? take 2° -2?-x. 

93, a?+l—-B8abc take b?—2abe. 

94, -8+4+6bc+b°c? take 4 —- 3be — 5b7c*. 

25, 3p*—2p°q + Tpq? take p*gq — 8pq?+q’. 
96. 7+x2-2 take 5-x+27+2°%. 

QT, —44+2°y-ayz take —3-2x°y+llaryz. 
98, —Sa’x?+5x?+15 take 9a?x? - 8x? -5. 
99, p®+r-38pqr take 1°+¢9°+ 3pqr. 

80, 1-32? take 2° -32?+1. 


“3, 2430-72? take 322- 32-2. 
82, x? +1la?+4 take 8a?- 52-3. 


83, a®+5-2a? take 8a*+3a? -7. 
04, 24+323-2?-8 take 244+ 32?-2#+2, 


v85, 1-2x4+32? take 7x? —42?4-32+1. 


36. xtyztyzx take — 3y?2u — Qary2? — xyz. 

37, 4a%a?-38aat+a® take 3a°x?+7a72x3 — a. 

88, l-w2+a°-at-az take zt-l+x-2". 

89, —-Smn?+15m2n+n3 take m3 —n?+8mn? —7m?n. 
40, 1-p? take 2p? - 3pq?- 2q'. 


33, The following exercise contains miscellaneous examples 
on the foregoing rules, 


IV. | SUBTRACTION. 19 


MISCELLANEOUS EXAMPLES I. 
1, When x=2, y=3, 2=4, find the value of the sum of 52’, 
-3xy, 2’. Also find the value of 327+ 324. 


9, Add together 3ab+be-—ca, -ab+ca, ab-2be+5ca. From 
the sum take 5ca+ be — ab. 


3, Subtract the sum of x-y+3z and -2y—2z from the sum of 
2a —-5y—3z and -3x+y+4+ 4. 


4, Simplify (1) 3b —- 2b? - (2b — 30%). 
(2) 8a-2b —(2b +a) —(a—5d). 
5, Subtract 8c?+8c-—2 from c®-1. 


6, When w=3, a=2, y =4, 2 =0, find the value of 
Arey 
9) Sata 
(2) 3y 
7, Add together 3a?-7a+5 and 2a*+5a-3, and diminish the 
result by 3a7+2. 


(1) 2a?- Bay +4az%, 





8, Subtract 267-2 from -—2b+46, and increase the result by 


S0— 1s 


9, Find the sum of 32?-47+8, 24-33-47, and 2z?-2, and 
subtract the result from 62743. 


10, What expression must be added to 5a?-3a+12 to produce 
9a?—7? 


11. Find the sum of 22, — 2°, 3x7, 2; +52, -4, 32°, ~—52?, 8; 
arrange the result in ascending powers of x. 


12, From what expression must the sum of 5a7-2, 8a+a?, and 
7 - 2a be subtracted to produce 8a?+a-—5? 


18, When x=6, find the numerical value of the sum of 1-2 +2", 
2a2—1, and x — x? 


14, Find the value of 6ax + (2by — cz) — (2au - 8by + 4cz) — (cz + a2), 
nauiema= 0,0 = 1,¢=2,0=8, y=3, 25 4. 


15, Subtract the sum of 4° — 3x7, 2a?-7a, 8a-—2, 5- 32°, 2a°-7 
from a3+a?+a+1, 


aA) ALGEBRA. (CHAP. IV. 
16, What expression must be taken from the sum of p*-—3p’, 
2p +8, 2p7, 2p? — 3p4, in order to produce 477-3? 


17. What is the result when -—32°+2z2-1lx2+5 is subtracted 
from zero. 
18, By how much does b+¢ exceed b-c? 


19, Find the algebraic sum of three times the square of x, twice 
the cube of x, —2?+a -22?, and 23-a—a?+1. 


20, Take p?-—q? from 3pq—4q", and add the remainder to the 
sum of 4pq — p? — 3q? and 2p* + 6q". 


91, Subtract 3b°+2b6?-8 from zero, and add the result to 
b4 — 2b? + 30. 


99. By how much does the sum of —~m?+2m-—1, m?-3m, 
2m? —2m?+5, 3m? + 4m?2+ 5m +8, fall short of 1lm* —8m?+3m? 

93, Find the sum of 82° -4a3y?, Taty—axy*, 3a°y?+ 2a°y3 + dary}, 
y—4ayt+a%y?, 2° —y+a3y24+ vyt- ay? +3aty, and arrange the 
result in descending powers of x. 


24, To what expression must 3x -4a3+7a7+4 be added so as to 
make zero? Give the answer in ascending powers of x. 


95, Subtract 7a?-32-6 from unity, and x - 52? from zero, and 
add the results. 





96, When a=4, b=3, c=2, d=0, find the value of 


(1) 8a? —2be —ad-+ 3b%ed. jp ee 
; Ja 
97, Find the sum of a, —3a?, 4a, —5a, 7, -18a, 4a7, —6, and 
arrange the result in descending powers of a. 


98, Add-together 44+327+23, 3-22-11, x?-—22?+7, and sub- 
tract 22°+2?—7 from the result. 


29, Ii a=5x-3y+2z, b= -2x+y-32z, c=x- 5y+6z, find the 
value of a+b-—-c. 


30, If «=2a?-5a+3, y= -8a?+a+8, z=5a?—6a-—5, find the 
value of «-(y+2). 


CHAPTER V. 


MULTIPLICATION. 


34, Mu.rreiicaTion in its primary sense signifies repeated 
addition. 
Thus 3x5=3 taken 5 times 
one Sone 
Here the multiplier contains 5 units, and the number of times 
we take 3 is the same as the number of units in 5 


Again axb=a taken b times 
=a+a+ta+a+..., the number of terms being 0. 
Also 3x5=5 x8; and so long as a and 6 denote positive whole 
numbers it is easy to shew that 
ax b=-b <a; 
Hence gota ~¢=(axb)xe=b xa x ce—bac 
Ux (aX C)=0 X Ox @ = 0c". 
Similarly we can show that the product of three positive 
integral quantities a, b, ¢ is the same in whatever order the 
factors are written. 


Un pene x SOX xox 0 = 2X 3 O50b = Gad, 


35. When the quantities to be multiplied together are not 
positive whole numbers, the definition of multiplication has to 
be modified. For example to multiply 3 by #, we perform on 
3 that operation which when performed on unity gives +; that 


is, we must divide 3 into 7 equal parts and take 4 of them. 


By taking multiplication in this sense, the statement ab=ba 
can be extended so as to include every case in which a and b 
stand for positive quantities. 


It follows as in the previous article that the product of 
number of positive factors is the same in whatever order the 
factors are written. 

386, Since, by definition, a®=aaa, and a®=aaaaa ; 

a? x @=aaa X aaaaa=aaaaqaaaa=a>= at ; 


that is, the index of a in the product is the sum of the indices 
of a in the factors of the product. 


Again, 5ba*=5aa, and 7a°=7aaa ; 
ba? x 7a=b x X aadaa= 35a", 


22 ALGEBRA. [ciaP. 


When the expressions to be multiplied together contain 
powers of different letters, a similar method is used. 


Example. 5a®b? x 8a*bx? = 5aaabb x Saabuxx 
= 40082) 


Note. The beginner must be careful to observe that in this pro- 
cess of multiplication the indices of one letter cannot combine in any 
way with those of ancther. Thus the expression 40a°b’z? admits of 
no further simplification. 


37. Rule. Zo multiply two simple expressions together, 
multiply the coefficients together and prefix their product to the 
product of the different letters, giving to each letter an index equal 
to the sum of the indices that letter has in the separate factors. 


The rule may be extended to cases where more than two 
expressions are to be multiplied together. 

Hxample 1. Find the product of 2”, x, and 2°. 

‘The product)= a7 xa SG20 =e xe ee, 

The product of three or more expressions is called the con- 
tinued product. 


Hxample 2. Find the continued product of 5x?y?, 8y°2, and 3x24. 


The product 2 Hay? Sy poe nl our ay ee. 


38, By definition, (a+b)m=m+m+me-+ ... takena+b times 
=(m+m+m-+... taken a times), 


together with (m+m+m-+... taken b times) 
=am+bm. 

Also (a—b)m=m+m+m-+ ... taken a—b times 
=(m+m+m-+... taken a times), 

diminished by’ (m+m+m+... taken b times) 
=am—bm. 

Similarly (a-—b+c)m=am—bm+em. 


Thus it appears that the product of a compound expression by 
a single factor ts the algebraic sum of the partial products of each 
term of the compound expression by that factor. 

Hxamples. 3(2a+3b- 4c) = 6a+9b —12c. 

(4a? — Ty — 82?) x Bay? = 12a3y? — Qlay® — Way, 


v.1 MULTIPLICATION. 29 


EXAMPLES V. a. 


Find the value of 


BL oa a BER Go Ar eee one GEOG a Od 
face < 7c 6, 9y? x 5y®. 7, 3m? xbm*, 8, 4a® x Gat. 
Q, 3xx 4y. 10. ba x 60: ieee oa; TO apt og". 
Domocaxdbax. 14, 3qrx4or. 15, abxab. 16, 3ac x 5ad. 
eee ate, 19. 3x°y*x4y2 19, abe xab. 90, atx da°b* 
Ql, a? x ab x 5ab4. OO Rie x Op rx [pr 
93, Ga®y x ay x 9aty?. OA 1G? <3b" x De, 
DD. Gay? x Tyz x az. 296, 3abcd x 5bca? x 4cabd. 
Multiply 
27, ab-—ac by are. 98, x*y—a®2+4y2 by a3y2?. 
99, 5a? - 3b? by 8ab?ct. 30, ab -—5ab+6a by 3a*d. 
Sigea-— 20? by 327. 32, 2ax?-b®y+3 by aray. 
83. Tp*a-part+l by 2p. 384, m?+5mn-3n? by 4m?n. 
35, zy? -3x?z-2 by 3yz. S00 oa Dy Zara. 
39, Since (a —b)m=am— bm, [Art. 38. | 


by putting c—d in the place of m, we have 
(a—b)(c -d)=a(c— d) —b(c—d) 
=(e—d)a—(ce—d)b 
=(ace—ad)-— (be— bd) 
=uac—ad — be+bd. 
If we consider each term on the right-hand side of this result, 
and the way in which it arises, we find that 
(+a)x (+e) =+ae. 
(—b)x(—d) = +d. 
(—6b)x(+c) =—be. 
(+a) x(-d)=-— ad, 
These results enable us to state what is known as the Rule 
of Signs in multiplication. 


Rule of Signs. Zhe product of two terms with like signs ts 
positive ; the product of two terms with unlike signs ts negative. 


24 ALGEBRA, [cHaP, 


40, The rule of signs, and especially the use of the negative 
multiplier, will probably present some difficulty to the beginner. 
Perhaps the following numerical instances may be useful in illus- 
trating the interpretation that may be given to multiplication 
by a negative quantity. 

To multiply 3 by —4 we must do to 3 what is done to unity 
to obtain — 4. Now —4 means that unity is taken 4 times and 
the result made negative: therefore 3x (—4) implies that 3 is 
to be taken 4 times and the product made negative. 


But 3 taken 4 times gives +12; 
oie 4) 

Similarly —3x —4 indicates that —3 is to be taken 4 times, 
and the sign changed ; the first operation gives —12, and the 
second +12. 

Thus (-—3)x(-4)=+4+12. 
Hence, multiplication by a negative quantity indicates that we are 
to proceed just as if the multiplier were posite, and then change 
the sign of the product. 


Hxample 1. Multiply 4a by —3b. 
By the rule of signs the product is negative ; also 4a x 3b = 12ab; 
*. da x (- 3b) = —12ab. 


Example 2. Multiply -5ab®x by - ab?a. 


Here the absolute value of the product is 5a7)°:*, and by the rule 
of signs the product is positive ; 


“. (—Sabix) x(—ab?x)= 5a7d's?. 


Example 3. Find the continued product of 3a7b, -2a3b?, — abt, 


This result, however, may be 


a2 WD 3h =o Aap hos i 
ie ae b m= oes written down at once : for 
Pe Obi )” x ia a = OL . F > 
3a7b x 2a30? x abt = 65’, 
Thus ee complete pro- and by the rule of signs the re- 
duct is 6a°b’. quired product is positive. 


EHxample 4. Multiply 6a? - 5a2b -4al? by - 3ab>. 


The product is the algebraical sum of the partial products formed 
according to the rule enunciated in Art. 37 ; 


thus (Ga? - 5a*b — 4ab?) x ( - 3ab?) = — 18a4h? + 15a7b3 + 12074, 


v.] MULTIPLICATION, 25 


EXAMPLES V. b. 


Multiply together 

eo = 2, 9. —3, 42. 3, —2*, -x. 4, —5m, 3m? 

5, —4¢q, 3q”. 6, —4y°, —4y°. 7, —3m3, 3m3. 8, 424, — 424, 
9, —3x, —4y. 10, —5a7,4~. 11, —3p?,-49°. 12, 3ab, —4abd. 

Popeoa, — 0°, Zab.» 14, —a, -—0,' =c?. 15, 3a, — 2b, —4c?, —d. 

16, —3ab, —4ac,3be. 17, -—2a?, -3a7b,-6. 18, —2p, -3¢q, 4s, -¢. 


Multiply 
19, -ab+ac-—be by -ab. 20, —38a*?-4ax+5z? by — ax’, 
91, ac—ac?+c* by —ae. 22, —2ab+cd -efby —32°y%, 


4], To further illustrate the use of the rule of signs, we add 
a few examples in substitution where some of the symbols 
denote negative quantities. 

Example 1. Tfia= —4, find the value of a’, 

Here a® = (-—4)? =(-4)x(-4)x(-4) = — 64. 

By repeated applications of the rule of signs it may easily be 


shown that any odd power of a negative quantity is negative, 
and any even power of a negative quantity is posztive. 


Example 2. Iia=-1, b=38,c= —2, find the value of — 3a‘bc3. 


Here -3atbe? = -3x(-1)*x 3x (- 2)? We write. down at 
= —~3x(+1)x38x(-8) once, (—1)*= +1, and 
= 72, (-2)? = -8. 


RENEE V..¢. 


‘al 3 “a WV 
faa —1,9b=0,°¢= 2,07 =, res find the value of 
rot: OF 0G. 3, an. 4, (-a)?. 
b, = 3c. 6, (-q)t tase 8, -ac. 
9, ad. 10, —acn. TES 3a", 12. 4(—¥- 
18,  2abe?. Thee Se. 15, -(a)* 16, -8a7q. 


17, —a'n?, 18, ac’. 19, —a%c?, 20, ?g. 


26 ALGEBRA. [cHAP. 


li ea 3 esha) fk 5; y = —], find the value of 
91, 3a-2y+4k. 99, -—4c—3x4+2y. 93, -4a+5y-x. 
94, ac-—ds3cy- yk. 95, 2ay -ka+4k?. 96, a? -2c?4+3y?. 
O17, -a-ayt3y". 98, ax-—yx-cy. 29, @-y-+y% 
80, a-2?-2y. Bl, c*y2-2act+ch% 82, acy—y*t+2a?. 


Multiplication of Compound Expressions. 


49. To find the product of a+b and e+d. 


From Art. 38, (¢1+b)m=am+bm ; 

replacing m by e+d, we have 
(a+b)(e+d)=a(e+d)+b(ce+d) 
=(c+d)a+(ce+d)b 
=ac+ad+be+ bd. 

Similarly it may be shown that 
(a—b\e+d)=ac+ad—be—bd; 
(a+b\e—d)=ac—ad+be—bd; 

(a—b)(e-d) =ac—ad — bc + bd. 
43, When one or both of the expressions to be multiplied 


together contain more than two terms a similar method may be 
used. For instance 
(a—b+ cm =am—bm+em ; 
replacing m by x—y, we have 
(a—b-+o\(e—y)=ale—y)—W(e-y) + (w—y) 
=(ax — ay) - (bx — by)+(cx —cy) 
=axn—ay — bx+by+cer—cy. 
44, The preceding results enable us to state the general rule 
for multiplying together any two compound expressions. 


Rule. Multiply each term of the first expression by each term 
of the second. When the terms multiplied together have like signs, 
prefix to the product the sign +, when unlike prefie — 3; the 
algebraical sum of the partial products so formed gives the com- 
plete product. 


45, It should be noticed that the product of a+b and x-y 
is briefly expressed by (a+6)(#—-y), in which the brackets 
indicate that the expression a+0 taken as a whole is to be 
multiplied by the expression v—y taken as a whole. By the 


v.] MULTIPLICATION. 27 
above rule, the value of the product is the algebraical sum of 
the partial products +ar, +bx, —ay, —by; the sign of each 
product being determined by the rule of signs. 


Hxample 1. 
The product 


Multiply «+8 by x+7. 
=(e-S)e4 7) 
=a 8a 72406 
= “7+ 152+56. 
The operation is more conveniently arranged as follows : 





aie. We begin on the left and work 
e+ 7 to the right, placing the second 
e+ 8x result one place to the right, so 
4+ Fa+56 that like terms may stand in the 
ae Fag Sree same vertical column. 
by addition, w7+15%+56. 
Hxample 2. Multiply 2a-3y by 4a-—7y. 
20 = DY 
4a —Ty 
8a? — 12xy 
— l4ay + 21y? 





by addition, 82? — 26ay + 21y?. 


EXAMPLES V. d. 
Find the product of 


tp as 7, a5. Q, 2-3, +4, 3, a-6,a-7. 

4, y-4, yt+4. 5h, 2+9, 2-8. 6, c-8, ct+8. 

Tak), k— 5. 8, m-—9, m+12. 9, w2-12,2+11. 
10.6 @=14, +1. 11 ep 10s p10. a Wine fe eer aro ae 
13, 2-4, —x+4. 14, -y+3, -y-3 15, -a+4,-a45. 
16, «2-10, -v7+8 17, -A+4, -k-7. 18, -y-7,-y-7. 
19, 2a-5, 8a42. 90, «2-7, 2a+5. Ole 3a 4 Oe 8: 
92, 3y-5, y+7. 23, 5m-4,7m-3. 94, Tp-2, 2p+7. 
95, x-—3a, 2a+3a. 26, 3a-2b, 2a+38b. 27, 5c44d,5c—4d. 
98, a-2x, 3at+2x. 29, Th+c, 7b-2e. 30, 2a-5c, 2a+4 5c. 
Ol.. 3a—by, 4a+y. 82,. 2y*-3z, 2y+3z. 388, xy+2b, xy—2b. 
04, 2u-3a,2x+3b. 85, 3x-4y, 2a+3b. 86, mn-—p, 2xy+3z. 


28 


46, 
KHxample 1. 


3x7 — 27 —5 





2x — 5 
62? — 4x”?-—10x 
— 15x”?+ 10% +25 





Ga 31942 +25. 


ALGEBRA. 


[ CHAP. 


We shall now give a few examples of greater difficulty. 


Find the product of 3a?-2a2-5 and 22-5, ~ 


Each term of the first expression is 
multiplied by 2x, the first term of the 
second expression; then each term of the 
first expression is multiplied by —5; like 
terms are placed in the same columns and 
the results added. 


Hxample 2. Multiply a-b+38c by a+ 20. 


aa 


b+ 3¢ 


& 2p : 
a?- ab+3ac 


2ab 


= 20° +6be 





a = ab + 38ac — 2b? + Bie 


AT, 


If the expressions are not arranged according to powers, 


ascending or descending, of some common letter, a rearrange- 
ment will be found convenient. 


Hxample. 
2a?— 3ab +4b? 
= Ha?+ Sab £40? 
— 10a4 + 15a*b -- 20a7b? 
+ 6a°b-— 9a?b?+ 12ab? 
8a2b? — 12ab? 





+ 1604 





— 10a4+ 21a) — 21a7b? 


+ 1664. 





Find the product of 2a7+ 4b? -3ab and 3ab —- 5a? + 4b". 


The re-arrangement is not 
necessary, but convenient, 
because it makes the collec- 
tion of like terms more 
easy. 


: EXAMPLES V. e. 


Multiply together 


1, 22-382-2, Qn-1. 

3. 2y7%-sy+1, 3y—1. 
Hh, 2a?-3a-6, a-2. 

7. 327-2447], 20-7. 
Q, xw?+x-2, w*-x4+2. 
ll, 2a?- 3a-6, a?-a+2. 
18, at+b-c, a-b+e. 


4a?-a -2, 2a+3. 
3a7+4e4+5, 4a —-5. 
5b? -2b4+3, -—2b-3. 
5c? -4c+8, -—2ce4+1. 
w= 24 +5, 27-2745. 
2k? —-3k=1, 3h? -—k=1. 


a-2b-—38c, a—2b+3c. 


v.] MULTIPLICATION, 29 


15, 2-xyty, w+ayt+y’. 16, @-2a%+227, a?+2aa +227. 
17, @-0?-38c?, -a?-b?-3c2. 18, w-3a?-a, 2?-3r+1. 
19, a®-6a+5, a’?+6a—-5. 90, 2yt-4y?+1, 2y4-4y?-1. 
91, 5m?+3-4m, 5-4m+3m. 22, 8a>-2a2-3a, 3a?+1- 5a. 
98, 2x2+2a?- 327, 87+24+2e% 94, a? +? -a7l?, a*b?- a+b. 
95, a +2°+8a274+8ara, a? + 3ax?- x? - 38a7n. 

96, 5Spt-p?+4p?-2p+3, p?-2p43. 

97, m®—2m*+3m3—4m?, 4m - 3m? + 2m. 

98, at+1+6a?-4a?-4da, a®-1+4+38a-3a?. 

99, a®+b?+c?+ab+ac—be, a-b-e. 

80, 214+ Gary? + y* — 4a%y—4ay®, — at — y* — Cay? — dary? — 4a°y. 


48, Although the result of multiplying together two binomial 
factors, such as +8 and w-7, can always be obtained by the 
methods already explained, it is of the utmost importance that 
the student should soon learn to write down the product rapidly 
by inspection. 

This is done by observing in what way the coefficients of the 
terms in the product arise, and noticing that they result from 
the combination of the numerical coefficients in the two bi- 
nomials which are multiplied together ; thus 

(e7+8)(@+7)=27+80+7x24+ 56 
=7+154+56. 

- (@ —8)(@—7) =a? — 8x —7# +56 
=x? —152-+56. 

(7+ 8)(4@—-7)=274+ 8x2 —Tx—56 


) =2'+x—56. 
(7 —8)'7+7)=x2°-—8x+ Tx —56 
=? -x2- 56. 


In each of these results we notice that ; 
1, The product consists of three terms. 


2. ‘The first term is the product of the first terms of the two 
binomial expressions. 

3. The third term is the product of the second terms of the 
two binomial expressions. 


4, The middle term has for its coefficient the sum of the 
numerical quantities (taken with their proper signs) in the second 
terms of the two binomial expressions. 


30 ALGEBRA. [oHar. v. 


The intermediate step in the work may be omitted, and the 
products written down at once, as in the following examples : 
(w7+2)\(v7+3)=2° + 5r+6. 
(v—3)(e+4)=a27+4-12. 
(v7+6)(@—9)=2?— 3x — 54. 
(a —4y)(x2 — 10y) =x? —14ay + 40y”. 
(7 —6y)(a@+ 4y) =x? — Qay — 24y?. 
By an easy extension of these principles we may write down 
the product of any two binomials. 
Thus (2% + 3y)(% — y) = 2a" + 3.2xy — 2Qxey — 3y? 
= 20° + xy — dy". 
(8a — 4y)(2a + y) =6x? — 8xy + Bry — 4y? 
= 6x? — Bay —4y’. 
(v7+4)(~—4)=n7 +427 —4x- 16 
=27°—16. 
(2 + 5y) (2x — 5y) = 40? + 10xy — 10xy — 257? 
= 44? — 25. 


EXAMPLES V. f. 


Write down the values of the following products : 


1, (a+3)(a—-2). 9, (a-7)(a—-6). 8, (w—-4)(~%+5). 
4, (b-6)(6+4). 5 Y-Hiy—D- 6, (a-1)(a-9). 
7, (c-5)(¢+4). 8. (x-9)(a— 3). OS (yA) ye7) 
10, (@-3)(a+8). ll, (%-5)(x— 8). 12, (a+7)(a—-7). 
13, (4-6)(/-6). 14, (a-5)(a+5) 15, (¢+7)(¢+7) 
16, (p+9)(p- 10). 17. (+5)(2=8). 18, (%-9)(%+9). 
19, (w-8a)(v+2a). 20, (a-2b)(a+2b). QI, (a-—4y)(x~—-4y). 
22, (a+4c)(a+4c). 23, (c-5d)(c-5d). 24, (p—2q)(p+2¢). 
95, (2e-8)(82+2). 26, (Ba-1)(2Q4+1). 97, (5x—2)(52+4+2). 
28, (8a+2a)(8a-2a), 29, (6u+a)(6x—-2a). 80, (7x+3y)(7x-y). 


CHAPTER VI. 


DIVISION. 


49, Tur object of division is to find out the quantity, called 
the quotient, by which the divisor must be multiplied so as to 
produce the dividend. 


Division is thus the inverse of multiplication. 


The above statement may be briefly written 

quotient x divisor = dividend, 

or dividend ~+ divisor = quotient. 
It is sometimes better to express this last result as a frac- 


tion ; thus 


dividend = quotient 
divisor : 


Example 1. Since the product of 4 and 2 is 42, it follows that 
when 4x is divided by x the quotient is 4, 


or otherwise, 4u—a = 4, 


Example 2, Divide 27a5 by 9a*. 





ma : 9705 2Taaaan We remove from the divisor 
The quotient = bt Oaae and dividend the factors com- 
mon to both, just as in arith- 
= 3a = 3a" metic. 
Therefore 27a’ + 9a? = 3a? 


Example 3. Divide 35a*b?c? by 7ab?c?. 
3d5aaa . bb. ce 
7a@. 6b. ce 


In each of these cases it should be noticed that the index of any 
letter in the quotient is the difference of the indices of that letter in 
the dividend and divisor. 


Cc 
= 5aa.c= 5a. 





The quotient = 


32 ALGEBRA. [CHAP, 


50. It is easy to prove that the rule of signs holds for 
division. 








Thus QOS eer 
a a 
a : a 
: —a —a 





Hence in division as well as multiplication 
like signs produce +, 
unlike signs produce —, 
Rule. To divide one simple expression by another : 


The index of each letter in the quotient vs obtained by subtracting 
the index of that letter in the divisor from that in the dividend. 


To the result so obtained prefix with its proper sign the quotient 
of the coefficient of the dividend by that of the divisor. 


Hxample 1. Divide 84a°x? by —- 12a4x. 


The quotient = ( — 7) x a'-4z3-1 Or at once mentally, 
= -— Tax? 84a°x? + (-12a4x) = — Jaz. 
Example 2. —45a%b?a*+(~ 9a8bx?) = 5a®ba?. 


Note. If we apply the rule to divide any power of a letter by the. 
same power of the letter we are led to a curious conclusion. 


Thus, by the rule Go? a? 9 = a; 
3 
but also roa J, 
a 
read be 


This result will appear somewhat strange to the beginner, but 
its full significance is explained in the Theory of Indices. 
[See Hlementary Algebra, Chap. xxxt.] 


Rule. Zo divide a compound expression by a single factor, 
divide each term separately by that fuctor, and take the algebraic 
sum of the partial quotients so obtained. 


This follows at once from Art. 38. 


Examples. (9x -12y+8z)+(-—3) = -—8x%+4y -2z. 
(86a°L? — 24a7b° — 20a4b?) + 4a*b = 9ab — 6b* — 5a°b, 


VI. ] DIVISION. 33 


EXAMPLES VI. a. 


Divide 

Peace by x. 2. Ga’ by 3a. 3, oa’ by at. 

4, 210 by 76°. 5, xy? by —axy. 6, —3axy*® by 3y. 
7, 4p?q? by -2pq. 8, 1l5m®n by -—5m. 9, —2m? by —ln. 


a0 tse" by —6a°. J], 35242 by -72. 19, ~7a*> by \—7. 
lo. —28p'¢ by 28p*, 14, —Ta* by —2’. 15/ 2dxy2? by -— 322, 
16, —12b%< by Gb?e® 17, -9k" by -k4. 18, 2h by —hi. 


19, —45a*b%cl’ by 9a7b?c!, 90) Saryte? by —ax%yz?. 
91. —168a7b?ca? by — Taba, 99, —35a%bSa? by — 7a*bta’. 
23. 327-2e by x. 94, 5a®b- Tab? by ab. 

25, 48p*q —24pq? by 8pq. 26, —152°+252* by —- 52°. 
O17, w-xy-—xz by -x. 98, 10a?-5a*b+a by —a. 


99, 403+36ax2—16«% by —4za. 30, 38a?—9a2b —Cab? by —3a. 


When the Divisor is a Compound Expression. 
51, Rule. 1. Arrange divisor and dividend in ascending or 
descending powers of some common letter. 
2. Divide the term on the left of the dividend by the term 
on the left of the divisor, and put the result in the quotient. 


3. Multiply the WHOLE divisor by this quotient, and put the 
product under the dividend. 


4, Subtract and bring down from the dividend as many terms 
as may be necessary. 


Repeat these operations till all the terms from the dividend are 
brought down. 
Example 1. Divide x?+11%+30 by «+6. 
Arrange the work thus : 
+6) 22+ 11x% +30 ( 


divide x", the first term of the dividend, by x, the first term of the 
divisor ; the quotient is x. Multiply the whole divisor by x, and put 
the product «7+ 6x under the dividend. We then have 


x+6)a*+1le+30(2# 
o+ 6x 


by subtraction, Ba +30 





On repeating the process above explained, we find that the next 
term in the quotient is +5. 


HA; Cc 


34 ALGEBRA. [CHAP. 


The entire operation is more compactly written as follows : 
x+6)x72+1l2+380(42+5 

a+ 6x 

ox + 30 

5x +30 








The reason for the rule is this: the dividend is separated 
into as many parts as may be convenient, and the complete 
quotient is found by taking the sum of all the partial quotients. 
By the above process #?+112+30 is separated into two parts, 
namely 27+ 6x, and 57+30, and each of these is divided by «+6; 
thus we obtain the partial quotients + and +5. 


Hxample 2. Divide 24a?- 65xy+2ly* by 8x -3y. 


Divide 2427 by 8x; this 


» — 297) D472 — B5y PTa2( 3x -7 
8a — dy) 24a" — Oday + 2ly"( 3x - Ty gives 32, the first term of the 





me ES) quotient. Multiply the whole 
- 56xy+21y? divisor by 38x, and place the 
— 56ay + 21y? result under the dividend. 








By subtraction we obtain 
—56xy+2ly?. Divide the first term of this by 8z, and so obtain 
—Ty, the second term of the quotient. 
Hxample 3. Divide 16a*—46a?+39a-9 by Sa-3. 
8a —3 ) 16a? — 46a? + 39a — 9 ( 2a? -5a+3 





l6a°— 6a? 
— 40a?+ 39a 
— 40a? + 15a 
94a —9 
Y%4a—9 





Thus the quotient is 2a? -5a+3. 


EXAMPLES VI. b. 


Divide 

], a+2a+1 by a+. Q, U?+3b4+2 by 6+2. 

3, 2?+42+3 by «+1. 4, y?+5y+6 by y+3. 

5, 27?+5x2-6 by x-l. 6, 2°+2x2-8 by 2-2. 

7, p?+38p-40 by pt8. 8. g@-4q-82 by q+4. 

9, a?+5a-50 by a+10. 10, m*+7m-78 by m-—-S. 
ll, #*+ax-30a* by x+6a. 12, a*+9ab—-36b? by a+120. 


vI.] DIVISION. 35 


Divide 
18, -—2?+18x%-45 by x-15. 14, «* 4274441 by x--21. 
15, 2au?-182-24 by 2443. 16, 5x74+16%+3 by «+3. 
17, 6%7+52-21 by 22-3. 18, 12a*+au-6xz? by 3a- 22. 


19, —52?+ay+6y? by ~x-y. 20, 6a? ac—35c? by 2a—-5e. 
21, 12p?--'749q+12q? by 2p-12¢. 

92, 4m?-49n? by 2m+7n. 

238, 12a7--3lab+20b? by 4a—-5bd. 

24, —2527+49y? by -5a+7y. 

25, 21p?+1lpq—40q? by 3p+5q. 

26, 82°+82?+4e+1 by 27+]. 

27, —2x3+1322-17x+10 by —x+5. 
98, x? +ax?-3a°x-—6u? by w-2a. 

29, Ga®y—a*y?-Txy?+12y* by 2a+4+3y. 
30, 8x?-122?-14%+21 by 2a-3., 


52. The process of Art. 51 is applicable to cases in which 
the divisor consists of more than two terms. 


Hxample 1. Divide a*t-—2a*-—7a?+8u+12 by a?-a-6. 
a*—-a-6)a*—2a?—7Ja?+8a+12(a2?-a-2 
a*—a?— 6a? 
-a-—a*+8a 
—a?+a"+6a 
— 2a7+2a+12 
— 2a7+2a+12 





Hxample 2. Divide 4x?-5x?+6x°-15-at-x by 3+22%--2, 
lirst arrange each of the expressions in descending powers of 2. 
2x7 -—2+3)62°— 244+ 403-5? -a2- 15 ( 822+ a?-2a-5 

6x? — 324+ 9x3 
24 —5a— 5a? 
se a ode 
—493— 827- x 
—4a3+ 2o7?- 6a 
*~ 1027+ 54-15 
—10x7+5x-15 








36 ALGEBRA. [CHAP. 


Hxample 3. Divide 23a7-2a4-4a3412+2>-3lx by x2?-7x+5. 


x? — Te +5 ) a° — Qat — 4a3 4+ 23a? — 31a +12 ( a? - 24+3 








xe” —TJao+ 5a 
~ 2x4 + 3a? + 18x? - 31a 
— 2x4 +14a?—-10x 
323+ 47-2174 12 
32° —21x%4+15 
 hege® ae 


Now 42? is not divisible by x’, so that the division cannot be 
carried on any further ; thus the quotient is «?-2x+3, and there is 
a remainder 4a7— 3. 

In all cases where the division is not exact, the work should be 
carried on until the highest power in the remainder is lower than 
that in the divisor. 


53. Occasionally it may be found convenient to arrange the 
expressions in ascending powers of some common letter. 


Example. Divide 2a?+10-16a-39a?+15a4 by 2-4a—-5a?. 
2-—4a-5a?)10-16a-39a?+ 2a°+ 15a4( 5+ 2a - 8a? 
10 — 20a — 25a? 
4a—14a?+ 2a° 
4a— 8a*-~10a° 
— 6a?+12a?+15a4 
— 6a?+12a*?+15a4 








EXAMPLES VI. c. 


Divide 

1, w-6a?+lla-6 by a?-4a+3. 

9, x -40?4+24+6 by x?-x-2. 

8. yty?-9yt12 by ¥°-3y+4+3. 

4, 2lm?-m?+m-1 by 7m?+2m+1. 

5, 6a*?-5a*-9a-2 by 2a7-3a-1. 

6, 66-k-14k4+3 by 3h°+4k-1. 

7, 6x?+1la?—-392-65 by 3x74 134413. 
8, 


122° — 8ax? — 27a7x%+ 18a? by 62? —13ax + 6a? 


vi.] DIVISION. 37 


Divide 
Q, 162°+14a?y —129xy? - 15y? by 8x7 + 27xy + 3y?. 
10, 2c? — 5c2d — 3ced? - 2d? by 7c?+3cd+d?. 
1], 324-1003 +12¢?-1l2+6 by 32°-2x+3. 
12, 30a*+1la’ —- 82a7-12a+48 by 8a7+2a-4. 
13, v°--%?-8x-13 by 274+32+4+3. 
14, a+38a?+6-10a? by a?-4a+3. 
15, 21m?-—27m-26m?+20 by 38m+7m?-4. 
16, 18x?+24a3 — 40a?x ~- 9ax? by 927+ 7a? - 18ax. 
17, 3y'-4y?+10y?+3y-2 by y-y7+3y+2. 
18, 5a?+1+10at- 4a? by 5a8-2a+1. 
19, l2a4*+5x3-332?-32+16 by 4a°-x-5. 
90, pt-6p?+18p?-10p+7 by p?-3p4+2. 
D1, 28x*+69x"+2— 71x? - 35x? by 40746 —- 13a. 
99, 5a°—7a4— 9a? —-1la?-388a+40 by -—5a?+17a—-10. 
93, «x°>-8a® by 2?+2axr+ 4a. 24, yi+9y?+81 by y?-3y4+9. 
95, «vt+4y* by x2 + 2ay + 2y". 96, 9at-—4a7+4 by 3a?-4a+2. 
D7, a®+64 by at-—4a?+8. 
98, 1624+ 362°4+81 by 4%7+62+9. 
99, 4m? — 29m -36 + 8:n?-—7m?+6m* by m? —2m?+3m -4. 
BO, 15a4 +22 — 320° - 302 +502? by 3-—4x + 52°. 
Bl, 3a?+8ab+4b?+10ac+8bce4+ 8c? by a+2b4+ 3c. 
62, 9u?-4y?+4yz2-2% by -—38x+2y—-z. 
63, 4c7-12c-—d?+9 by 2c+d-3. 
84, 9p? -16q?+380p +25 by -38p-—4q-5. 
O06 Ce y tery —e+2°—y by -x-y. 
86, xv +a2ty—2%y?+23-Qay? by 22 +ay—-y?. 
37, a°b?+ab—9 —b4+ 3b? + 3b - at -— 3a? - 3a by 3-b+a’°, 
88, z4+1 by v4+a?+e4+1. 39, 2a°+2 by a?+2a?+2a+1. 
40, 2°-—6a4-82°-1 by 2? -2e-1. 


CHAPTER VII. 


REMOVAL AND INSERTION OF BRACKETS. 


54, Quantities are sometimes enclosed within brackets to 
indicate that they must all be operated upon in the same way. 
Thus in the expression 2a--3b—(4a—2b) the brackets indicate 
that the expression 4a—26 treated as a whole has to be sub- 
tracted from 2a — 30. 

It will be convenient here to quote the rules for removing 
brackets which have already been given in Arts. 24 and 25. 

When an expression within brackets is preceded by the sign +, 
the brackets can be removed without making any change in the 
CXPTeSSON. 

When an expression within brackets is preceded by the sign —, 
the brackets may be removed if the sign of every term within the 
brackets be changed. 


Example. Simplify, by removing brackets, the expression 
(2a — 3b) — (3a + 4b) — (b - 2a). 
The expression = 2a — 3b —- 38a-4b-—b+2a 
= a-8b, by collecting like terms. 

55, Sometimes it is convenient to enclose within brackets 
part of an expression already enclosed within brackets. For 
this purpose it is usual to employ brackets of different forms. 
The brackets in common use are ( ), { }, [ ]. 

56, When there are two or more pairs of brackets to be 
removed, it is generally best to begin with the innermost pair. 
In dealing with each pair in succession we apply the rules quoted 
above. 

Hxample. Simplify, by removing brackets, the expression 

a — 2b — [4a — 6b - {3a -c + (2a — 4b +¢)}]. 

Removing the brackets one by one, 

the expression = a ~ 2b —- [4a - 6b — {3a —c+2a -—4b+c}] 
= a—2b—[4a - 6b-38a+c¢-2a+4b -c] 
=a—-2b-4a+6b+38a—c+2a-4b+¢ 
=2a, by collecting like terms. 


Note. At first the beginner will find it best not to collect terms 
until all the brackets have been removed. 


CHAP. VII.] REMOVAL OF BRACKETS. 39 


EXAMPLES VII. a. 


Simplify by removing brackets and collecting like terms : 


1, @+2b+ (2a - 3b). 9, a+2b-(2a-3b). 

8, 2a-—3b—-(2a+25). 4, a-2-(4-8a). 

5, (x—3y) + (2u—-4y) —(x-8y). 6, @+2b-38¢-(b-a-4e). 
7, («-8y+2z) -(2-4y+4+2z). 8, 4a -(2y +22) — (38x - dy). 
9. 2a+(b—3a) — (4a — 8b) — (6b — 5a). 


10, m-(n-p)—(2m-2p+28n) -(n-m+2p). 
1], a-b+c-(a+c-—b)-(a+b+c) -(b+e-a). 
12, 5a—-(Ty+3x) — (2y+7x) — (8a + 8y). 

13, (p-49)-(¢-2p)+(2p — 9g) — (p - 29). 

14, 2a? - (3y?— 22) — (a? 4y?). 

15, (m?— 2n?) — (2n? — 3m?) — (3m? - 4n?), 

16, (#—-2a) —(x% — 2b) —{2a—a%-(2b+2)}. 

17, (a+3b) -(b—3a) —{a+2b -(2a-3)}. 

bee 9 (9° + 2p") —{p" £39? — (2p? — 9’)}. 
19, x-[y+{e-(y-x)}} 20, (a—b)-{a-b- (a+b) -(a—D)}. 
21, p-[p-(¢+p)-{p—(2p-4)}]. 

92, 8e-y-[w-(2y-2)-{2e—(y—2)}} 

93, 38a? —[6a? — {8b? — (9c? — 2a7)}]. 

Q4, [Ba — {2a —-(a—b)}] - [4a — {3a — (2a — b)}}. 


57, <A coefficient placed before any bracket indicates that 
every term of the expression within the bracket is to be multi- 
plied by that coefficient ; but when there are two or more 
brackets to be considered, a prefixed coefficient must be used as 
a multiplier only when its own bracket is being removed. 


Examples 1. 244+3(4—-4)=2x44+3%-12=527—-12. 
me 2. Ta-2(x-—4)=7x-2~4+8=527+4+8, 


Lxeample 3. Simplify 5a —-4[10a+ 3{z-a-2(a4+2x)}]. 


The expression 


=5a—4[10a+3{x-a~2a-2zx}] On removing the innermost 
= 5a —4[10a+ 3{ —2—3a}] bracket each term is multiplied 
=5a —4[10a - 32—9a] by -2. Then before multiply- 
ing by 3, the expression within 
=5a—4[a - 32] its bracket is simplified. The 
=5a—4a+12e ‘ | other steps will be easily seen. 


=a+12z, 


40 ALGEBRA. [oHar. 


58. Sometimes a line called a vinculum is drawn over the 
symbols to be connected ; thus a—b+c is used with the same 
meaning as a—(b+c), and hence a—b+c=a-—b-c. 


Nort. The line between the numerator and denominator of a 





fraction is a kind of vinculum. Thus “ 3 5 ig equivalent to +(”—5). 
Laxample 4. Find the value of 
84 —7[ —lla-—4{ -17%+3(8 -9- 5x)}). 
The expression = 84 —7[ - llw —4{-17x+38(8 -9+5z)}] 
= 84-7[-lla-—4{-17x%+3(5x—-1)}] 
= 84-7[ -llw-4{-17x+4+ 15x -3}] 
= 84-—7[-lla—4{— 2a --3}] 
= 84—-"7[-lla+S8x+12] 
= 84-—{4[—32+4+ 12] 
= 84+2la - 84 
= 21x. 
When the beginner has had a little practice the number of steps 
may be considerably diminished. 


Insertion of Brackets. 

59, The rules for insertion of brackets are the converse 
of those given on page 12,and may be easily deduced from them. 

For the following equivalents have been established -in 
Arts, 24 and 25: 

a+b—c=a+(b—c), 
a—b—c=a-—(b+e), 
a—b+c=a-(b-c). 

From these results the rules follow. 

Rule, 1. <Any part of an expression may be enclosed within 
brackets and the sign + prefixed, the sign of every term within the 
brackets remaining unaltered. 

LHxamoples. a—-b+c-d-e=a-—b+(c-d-e). 

a? —aue+ ba —ab = (x? - ax) + (ba - ab). 

Rule, 2. <Any part of an expression may be enclosed within 
brackets and the sign — prefixed, provided the sign of every term 
within the brackets be changed. 

Examples. a-b+c-—d-e=a-(b-c)-(d+e). 

xy —ax—by+ab = (xy — ly) —(ax-ab). 


vir. ] REMOVAL AND INSERTION OF BRACKETS. 41 


60, The terms of an expression can be bracketed in various 
ways. 


Example. The expression aw - bu+cx—ay+by -cy 


may be written (aa — bx) + (cx — ay) + (by — cy), 
or (aw - ba +cx) -(ay —by+cy), 
or (ax — ay) - (ba — by) + (cx - cy). 


61, When every term of an expression is divisible by a com- 
mon factor, the expression may be simplified by dividing each 
term by this factor, and enclosing the quotient within brackets, 
the common factor being placed outside as a coefficient. 

Thus 30 —21=3(47—-7) ; 
and x? — Qax + 4a? =x? — 2a(x— 2a). 


EXAMPLES VII. b. 
Simplify by removing brackets : 
1, 3(x%—-2y) —2(a - 4y). 2, x-3(y-x)-4(x-2y). 
8, 16-3(2%-3)-(2x¢+8). 4, 4(~+3)-2(7+a)4+2. 
5, 8(a-—3)-(6- 2x) -2(a4+2)4+5(5- 2). 
6, 2u¢-5(38a—7+y)+4(2x + 3y —- 8) - Ty. 
7, 2u-—5{8x—-7(4e-9)}. 
8, 2 8ay se) tP — 0 —3(x°y — ay") - y°. 
9, 4a —3{x-(1-y)+2(1-2)}. 
10, x-(y ee Cen 2) 
Pie = (ete a 9") — 297} 1 y"), 
12, 5x+4(y- 2z)—4{a+2(y—-2)}. 
13, a+{—2b+3(c-d-e)}. 
14, {a?- (b? -c?)} - [2a? — {a — (b? — c?)} — 2(b? - c?)], 
15, 38p- {dq ~ [6g + 2(10q — p)]}. 
16,, 3x -—2[22 —{2(a—y)- i - y}. 
17, 3(5- 6x) -5[a—5{1 — 3(x—5)}}. 
omit (Gq — (7 — ea — {5a+(3-2- ae 
19, 0?— {a?+ab—(a*+b")} - [a? - {8ab — (b? - a*)}). 
90, cs In, ee eile at 
91, 20(2—x)+3(u-7) - 2a +9 3{9 -4(2-a)}}. 
99, -—A4(aty)+24(b - a) -—2[e+y+a—3ly+a-4(b+x)}), 





42 ALGEBRA. (cHaP. 


93. Multiply 
2x -3y —4(x -2y)+ 5{3x -2(x- y)} 
by 4x-—(y-—2)-3{2y-3(x+y)}. 
Jn each of the following expressions bracket the powers of x so 
that the signs before the brackets may be (1) positive, 
(2) negative. 
94, aavt+22° -—cx? +22? -b2e— at. 


2 


95, ax*+a22? — bx? —527- cz. 


MISCELLANEOUS EXAMPLES II. 


], Find the sum of a—2b+c, 3b-(a—c), 3a—6+3e. 

9, Subtract 1-«* from 1, and add the result to 2y — 2”. 

3. Simplify a+2b-—3¢+(b-3a+2c) —(3b-2a—-2e). 

4, Find the continued product of 3z*y, 2ry", —Tary®, —5aty’. 

5, What quantity must be added to p+q to make 2g? And 
what must be added to p* -3pq to make p*+2pq+q"? 

6. Divide 1-62*+5z° by 1-x+327. 


7. Multiply 3b?+2a?-5ab by 2a+3b. - 
8. When x =2, find the value of 1 -x+27- i< : 


= as 





9, Find the algebraic sum of 3az, —2zz, ax, —Txz, 4ax, —42z. 
10. Simplify _ 9a — (2b -—c)+2d - (5a+3b)+4c-2d, and find its 
value when a=7, b=~—3, c=-4. 
11, Subtract az*—4 from nothing, and add the difference to the 
sum of 22° — dz and unity. 
12. Multiply 32°y-—42y2+22y*2 by -6z°y*2 and divide the 
result by 3xy7z". 


18, Simplify by removing brackets 5[2-—4{x-3(2xr-—3xz+2)}]. 

14, Simplify 2a?- (2ay - 3y*) + 4y? + (Sry — 2z*) + a? — (Qry + Gy”). 

15, Find the product of 2x—7y and 3x+8y, and multiply the 
result by «+ 2y. 

16. Find the sum of 34426, -5ce-2d, 3e€+5f, b-a+2d, 
—2a -3b+ 5¢— 2. 

17. Divide x*—42°-18z*-11lz+2 by z?-7r+l. 

18, If a=-1, b=2, c=0, d=], find the value of 

ad+ac—a®-cd+e2—a+2c+a%b +2a3, 


Fira Fs 


vil.) 


19. 


MISCELLANEOUS EXAMPLES II. 435 


Simplify 3[1 —2{1-—4(1 —3z)}], and find what quantity must 
8z. 


be added to it to produce 3-— 


20. 


Divide the sum of 102°-7z(1+z*) and 3(z*+2z°+2) by 


3{a*+1)-—(z+1). 

91, Simplify 5z*— 827 — (227-7) —(z*+5) + (32° — x), and subtract 
the result from 4z*-z+2. 

99. Ii a=0, b=1, c=3, d=-2, e=2, find the value of 


23. 


(1) 3e°—d*5 (2) (ctaYle—a)+h5 (3) e +a. 
Find the product of 7z* — y{z—2y) and z{7z+y) — 277. 


94. Subtract (a2°+4)+(a*- 2) from (a@°+4)(a°-2 


25. 


saat 


27. 
28. 


29. 


30. 


Express by means of symbols 
(1) 5's excess over c is greater than a by 7. 
(2) Three times the sum of a and 2 is less by 5 than the 
product of 5 and ¢. 
Simplify 
3a? — (4a — 5°) — {2a? — (3b — a*) — 2b — 3a} — {56 —Ta —(C2-B)}. 
Find the continued product of 
@+tayty, 2-ayty, -xyPty* 
Divide 4a*-—9F—4ac+e by 2a-—35-c¢. 
lf a=3, b=-2, c=0, d=2, find the value of 
(1) e(aib)+b(at+e)ta(e—b); (2) at+dé. 
From a rod a+5 inches long b-—c¢ inches are cut off; how 


much remains? 


SL. 


A boy buys a marbles, wins 5, and loses ¢; how many has 


he then? 


32. 


33. 
34. 


Simplify 24 —{5a—[Sa—(25+a)]}, and find the value of 
(a —b)a?+ b(a+5)] when a=1, 5=2. 

Divide 1 —5a2*+4¢° by 2°-2r+1. 

Multiply the sum of 32°—Szy and 2ry—y° by the excess of 


Sa-+9° over 27 +3zry. 


30. 


36. 


Express in algebraical symbols 
(1) Three times z diminished by the sum of y and twice z. 
(2) Seven times a taken from three times 6 is equal to five 
times the product of ¢ and d. 


(3) The sum of m and n multiplied by their difference is 
equal to the difference of the squares of m and n. 


ia=2. 6=1, <=0, d=~—1, find the value of 
(d- b)\(c — b) + (ac— bd)? +(e—d)(2e-4). 





CHAPTER VIII. 


REVISION OF ELEMENTARY RULES. 


[If preferred, this chapter may be postponed until the chapters 
on Simple Equations and Problems have been read. } 


Substitutions, 


62, Derixition. The square root of any proposed expression 
is that quantity whose square, or second power, is equal to the 
given expression. Thus the square root of 81 is 9, because 9°=81. 

The square root of a is denoted by </a, or more simply ,/a. 


Similarly the cube, fourth, fifth, &c., root of any expression 
is that quantity whose third, fourth, fifth, &c., power is equal to 
the given expression. 


The roots are denoted by the Symbols sa), x/s ais We. 
Examples. %/27=33 because 3° = 27, 
Rave 23 because 2° = 32. 


The root symbol ,/ is also called the radical sign. 


Example 1. Find the value of 5,/(6a®b‘c), when a=3, b=1, c=8. 
5, /(6e°b4c) =5 x (6x3? x 14x 8) 
= 5x ,/(6« 27 x 8) 
= 5x (3 x27) x (2x8) 
= 5x0 X4 
= 130). 
Note. Anexpression of the form ,/(6a%b4c) is often written V6a5b4c, 


the line above being used as a vinculum indicating the square root 
of the expression taken as a whole. 





Hxample 2. Ifa=-4,b= —38,c= —1, f=0, x =4, find the value of — 
TN (a2cex) ~ 3 S042 +5 /(f22). 
The expression =7 ai (= 4)* —1)42 3. = 9/4 1)724-0 
=7 2/(-64)-3,/81 
=7x(-4)-3x9 
=— 50d. 





CHAP. VIII.] REVISION OF ELEMENTARY RULES. 45 


EXAMPLES VIII. a. 
Ia=40=1, ¢=6, ¢d=0, find the value of 


1. Jb. 2, /9ab. 3. 60%. 4, J/9a*b?, 
5, /4b%e2. 6. 6a4l ec. 7, av9ac. 8, 3b 2/Bar?. 
9, Jaeb?—/9e2. 10, 3M abcd? — d V2a%b +/6ac. 


If a=-3, b=2, c=-1, x=-4, y=0, find the value of 
ll, Verzex. 12, »/3ac?. 13, 6abz. 14, 5/cx. 
15, V3abex. 16, rade. 17, dex. 18, %/3a2b%c. 


19, (Bac —J cx +n/b2cx. 20. Ney +nr/20% -/ 9a. 
Ol. Lk @=100, y=81, 2=16, find the value of 


Wee Vy t+ N4z. 
vo ee li a=—6, b=2, c=—1, x=—4, y=0, find the value of 


2n/ ae — 2n/ a2bta5y5 + /8a2d. 








Fractional Coefficients and Indices. 


63. Fractional Coefficients. The rules which have been 
already explained in the case of integra] coefficients are still 
applicable when the coefficients are fractional. 


Hxzample 1. Find the sum of 32?+ixy-—fy?, —2?-2ay+2y?, 
2 an2 54/2 
gv" — wy —4y". 


ge" + gay — ay? 
Here each column is added up separately, 





22 2 
— x? -—fxy+2y : : y 
Gere ie and the fractional coefficients combined 
3 aE: by the rules of arithmetic. 
12 4 1,,2 
ger — Bey toy 


Example 2. Divide 42°+ Ayay?+7ey? by tx+H4y. 
1 1 2 2 5 1 
e+ 3y ) ae tery? + ey? ( ga" — fay + ty? 

Ber + § xy 

~ EY tpory” 
1 2 1 2 
~@vy— ort 
9 

gay? + asy° 

244 
3 xy + ty? 





46 ALGEBRA. (CHAP. 


64, Fractional Indices. In all the examples hitherto ex- 
plained the indices have been integers, but expressions involving 
fractional and negative indices such as a3, #74, 3ct+a7t-Q, 
a-?—4a~'v —3v? may be dealt with by the same rules. For 
a complete discussion of the theory of Indices the student is 
referred to the Hlementary Algebra, Chap. xxxt. It will be 
sufficient here to point out that the rules for combination of 
indices in multiplication and division given in Chapters v. and vL 
are universally true. 





2 3 2 


“ we 3 wy 
Hxample\. «* xx = gi T= yl, 
Haample2. “a>*xat=a7*4=aq%=1. [See Note; Art: 50.) 


ab or ae pCa Coes el = <S5i 
Taample 3.247019 3G) 70 = 60220 6 2 = 00 6 ee Dom 


Die ae ae 2-3 3-5 -1 4 
Heample4., 34-4" 2 4" Se sys? ore ey, 
=204), 27 <2 -2-2,241 -4,5 
Heampled m0, 0" 2.0 6 A= Cee DF ae 


It will be seen from these illustrations that the rules for 
combining indices in multiplication and division may be con- 
cisely expressed by the two statements, 

(1) a” x eee (2) a= a" ia: 
where m and » may have any values positive or negative, 
integral or fractional. 


65, We shall now give some examples involving compound 
expressions. 





Example 1. Multiply et — 30844 by Qe — 1, 
as ~ 32° + 4 
Qu* -l 
22 — Gx? : Sxt 
- xe ce 3a2 —4 





a) 


"4 


Qe — Tx + la? —4 
Example 2, Multiply cx+2c-*-7 by 5-38c7*%+2cx. 


eo Tree * ' 

Oot 1580-2 Here the expressions have 

POSTE OORT Wee i Pe ae been arranged in descending 

2c**-14c%+ 4 powers of c, and it should be 
+ 5ce®—354+10c-* noticed that in this arrange- 


2 yeaah 19 2 ment the numerical terms —7 
Me eet ee OC eg and +5 stand between the 
2c%*— Yc*—-34+31c-*-6c-** terms involving c* and ¢c~*. 


vill. | REVISION OF ELEMENTARY RULES. 


Example 3. Divide 


1 


1 ee 2 5 Ley 3 
24a*-1l6x *++a*-16x *-5ax* by 8x Ont gt 


Arrange divisor and dividend in descending powers of x. 





5 3 1 
=o 46" 1 Out® 














308+ 4o*+160*-~ 16a 
~ 8x4 Gxt + 12a — 24a 
~ Qet 4 dat + Set - 1627 
?_ 16a" 


3 a = 
— Q2u?+ 4x4*+ 8x 


EXAMPLES VIII. b. 


3 7% sy Be 5 1, a By ee 
at —Qxr* —4u4*+82% a) he kee 04s ee 4_ 162 Piel 
7 


47 


1 
444, 
ne 
3-22 2 
as 
4 
3 
4 











], Find the sum of —im-—in, -2m+2n, —2m- 

2, Add together 2a-1b+1c, da—4b, ta+7~yb+10e, - es ab — de, 

8. From a+4b-—4c take ta—-b+he. 

4, Subtract 4a?+4ab—1b? from ta?-1ab+416?. 

5, Multiply 4a7+4y? by 4du- ty. 

6, Find the product of 4a?-3e+1 and dv+4, 

7, Divide 2a°-3y? by 4a -4y. 

8, Divide a?-—2a7b+4,ab?—- 2b? by a?-$ab+2b7. 

9, Simplify 1(2a-3y) - 1(3a-+2) 2y) + ay (7x — By). 
10, Find the sum of 

4y Ta +7Y—h a tetany, —3y-ytte 
ll, Find the product of 
da —1y+He-dy) and Ha-z)-Hy-4). 
12. Simplify by removing brackets 
a bd b 
ag 9) +5{24-a(a-3) } 

18, Divide 4a°-1ia?+1xa-4 by 3u-3. 
14, Subtract ~.(7z- °y) from 4(x% - 3y) -— Fy — 22). 
15, Add together (~— jy)(fx+y) and (2x -}y)(4u—y). 
16. Mutitiply 2a?-4a7~7+ ee 8q,—2x, 
17, Divide 36a?+1b?+14-4ab-6a+1b by 6a-14b- 


18, Simplify 6{x ~ 8(y -$)} (422 -y) + 2y-1)} 


48 ALGEBRA. [cITAP. 


19, Multiply $a?-3ab+b? by a?+4ab—2b?, and verify the result 
when a=1, b=2. 


<)fary 


90. Multiply x—- ty? +y by « Be 


oy) 


91, Divide eyes ypty by ae yt +y!. 
92, Find the product of ety+ys and a® ys q 
93. Multiply ae by AS age 
94, Divide c-3-8c-1-8 by c-1-3. 
95, Divide 
$,j-2 pe 5 -% -2, Le elk 

AoeBy-2—12x3y-1425 - Qa Sy 4+16a 8y? by 2a8y-1-344a By, 

96, Find the value of (av-2+a~1x)(ax-?-3a-12). 
ee 

27, Find the square of at =| aes. 


1 
98, Find the continued cpa OL 30.4b, a, 02" — bs and ae +b. 
Ek 
ei, 


99, Divide x-y by byt 4? yr saty 
80, Multiply a?+2a-2-7 by 5+a?- 
81, Find the value of cau e —x~%y-}), 


BN 
2) 
a2 


Important Cases in Division. 
66. The following example in division is worthy of notice. 
Example. Divide a’+b?+c?—38abe by a+b+e. 


a+b+c)a*—8abe+ 0+? (a?—ab-ac+b?-bc+c? 
a?+ a?b+arc 
— a*b—ac—3abe 
-— a*b—ab?—- abe 
—a*c+ ub? —2abe 

















—arec = abe —ac? 
ab?— abe+ac?+b? 
ab? +03 +b%e 
— abc+ac?— lec 
—~ abc — b?c — be? 
ac+be+ 3 
ae? + be? gio Ce 


Here the work is arranged in descending powers of a, and the 
other letters are taken alphabetically ; thus in the first remainder 
a’b precedes a*c, and ac precedes 3abe. A similar arrangement will 
be observed throughout the work. 


vill. | REVISION OF ELEMENTARY RULES. 49 


67. The following examples in division may be easily veri- 
fied; they are of great importance and should be carefully 
noticed. 


2 22 
OV Hat+y, 


eo 
O-yP_ 9 9 
I. aad +ryt+y, 
Ota aig 2 aye 3 
= Tey ty + y, 


and so on; the divisor being v—y, the terms in the quotient all 
positive, and the index in the dividend either odd or even. 





Py 9 9 

ra = LE — LY TY", 

Aa: 

ae = 98 — ody + ady?— xdy8-4+o2yt— ayy, * 


and so on; the divisor being «+y, the terms in the quotient 
alternately positive and negative, and the index in the dividend 
always odd. 





ia eae 
cael 
4. 4 
Th aes See ee ae 38 
I ASOT L?— HY + ity” —Y", 


em) ee 5 fre vot 35/2 — 9229/3 4 __ 4/5 

4 = 0 — HY + PY? — “ry? + LY* — YY”, 

LY 
and so on; the divisor being x+y, the terms in the quotient 
alternately positive and negative, and the index in the dividend 
always even. 


IV. The expressions 2?+y?, at+y', «+y%,... (where the 
index is even, and the terms both positive) are newer cxactly 
divisible by #+y or by x-y. 

All these different cases may be more concisely stated as 
follows : 

(1) w”—y" is divisible by «—y if n be any whole number. 

(2) «+7 is divisible by &+y if 2 be any odd whole number. 

(3) «—y" is divisible by x+y if n be any even whole number. 

(4) «"+y” is never divisible by «+y or by x—y, when n is 
an even whole number. 

H.A. D 


50 ALGEBRA. [cHapP. 


Dimension and Degree. 


68, Each of the letters composing a term is called a dimen- 
sion of the term, and the number of letters involved is called 
the degree of the term. Thus the product abe is said to be of 
three dimensions, or of the third degree; and ax* is said to be of 
five dimensions, or of the fifth degree. 

A numerical coefficient is not counted. Thus 8a20' and a20® 
are each of seven dimensions. 


69, The degree of an expression is the degree of the term 
of highest dimensions contained in it; thus a!—8a3+3a—5 is 
an expression of the fourth degree, and aa —'7b°x* is an expression 
of the fifth degree. But it is sometimes useful to speak of the 
dimensions of an expression with regard to some one of the 
letters it involves. For instance the expression ax? — bx?+cx—d 
is said to be of three dimensions in x. 


70. A compound expression is said to be homogeneous when 
all its terms are of the same degree. Thus 8a6—a4h?+9ab® is a 
homogeneous expression of the sixth degree. 


It is useful to notice that the product of two homogeneous 
expressions is also homogeneous. ; 

Thus by Art. 47, 
(2a? — 8ab + 4b") ( — 5a?+3ab+ 4b?) = —10a*+ 21a%b — 21a7b? + 165+ 

Here the product of two homogeneous expressions each of 
two dimensions is a homogeneous expression of four dimensions. 


Also the quotient of one homogeneous expression by another 
homogeneous expression is itself homogeneous. 

For instance in the example of Art. 66 it may be noticed 
that the divisor is homogeneous of one dimension, the dividend 
is homogeneous of three dimensions, and the quotient is homo- 
geneous of two dimensions. 


EXAMPLES VIII. ec. 


1, Divide a*+30ab—-1250°7+8 by a-5b+2. 
2, Divide #°+y2-2+38ayz by x+y-z. 

8. Divide a®—b?+14+38ab by a-b+1. 

4, Divide 18cd+1+27c?-—8d* by 1+38c-—2d. 


Vill. } ' REVISION OF ELEMENTARY RULES. 51 


. e * . ° e . . 
Without actual division write down the quotients in the 
following cases : 














x—1 a? + 5% a4 —at a*— oF 

5, = t G.. ath Tt es Oe 8. eta : 
1+a? 16-64 a+ 5° a — b° 

9, rea 10. 24d Il. atb~ 12, Ou 
a + 27y8 af — x6 +1 a6 + y8 

i “e+ By | 14, a+x « 15, c+ ios 16. a2 fy 


17, In the expression 
2a°b? + 8ab4* + 3a7b°x — 2° + 20a2b? — llat+ 7a%b?, 
which terms are /ike, and which are homogeneous ? 


18, In each term of the expression 
Ta*be? — ab?c + 12b%c4 — bec, 
introduce some power of a which will make the whole expression 
homogeneous of the eighth degree. 


19, By considering the dimensions of the product, correct the 
following statement 


(8a? — 5ay + y*)(8a? — Qay — 3y") = Wdat - 46274 + 9x?y? + 18xy? - 3y°, 
it being known that there is no mistake in the coefficients, 


90. Write down the square of 3a?-—2ab-—6?, having given that 
the coefficients of the terms taken in descending powers of a 
are 9, —12, —2, 4, 1. 


91, Write down the value of the product of 3a7b+5a?-— ab? and 
ab? + 5a? — 3a*b, having given that the coefficients of the terms when 
arranged in ascending powers of b are 25, 0, —9, 6, —1. 


92, The quotient of 2?-y?-1-3zy by x-y-1 is 
we +ay+at+y?—y+l. 
Introduce the letter z into dividend, divisor, and quotignt so as to 


make them respectively homogeneous expressions of three, one, and 
two dimensions. 


NIVERSITY Or | 
pesNors LIBRARY 


CHAPTER IX. 


SIMPLE EQUATIONS. 
71. AN equation asserts that two expressions are equal, but 
we do not usually employ the word equation in so wide a sense. 


Thus the statement 7+3+2+4=2¢7+7, which is always 
true whatever value x may have, is called an identical equation, 
or briefly an identity, 

The parts of an equation to the right and left of the sign 
of equality are called members or sides of the equation, and 
are distinguished as the right stde and left side. 


72, Certain equations are only true for particular values of 
the symbols employed. Thus 3v=6 is only true when #=2, 
and is called an equation of condition, or more usually an 
equation. Consequently an identity is an equation which is 
always true whatever be the values of the symbols involved; 
whereas an equation (in the ordinary use of the word) is only 
true for particular values of the symbols. In the above example 
3v=6, the value 2 is said to satisfy the equation. The object 
of the present chapter is to explain how to treat an equation of 
the simplest kind in order to discover the value which satisfies it. 


73, The letter whose value it is required to find is called 
the unknown quantity, The process of finding its value is 
called solving the equation, The value so found is called the 
root or the solution of the equation. 


74, An equation which involves the unknown quantity in 
the first degree is called a simple equation. It is usual to 
denote the unknown quantity by the letter 2. 


The process of solving a simple equation depends only 
upon the following axioms: 


1. If to equals we add equals the sums are equal. 

If from equals we take equals the remainders are equal. 
Tf equals are multiplied by equals the products are equal. 
If equals are divided by equals the quotients are equal. 


ge bo 


CHAP ~.] SIMPLE EQUATIONS. 


Oot 
Co 


75, Consider the equation 7z=14. 

It is required to find what numerical value x must have to 
satisfy this equation. 

Dividing both sides by 7 we get 


io, [Axiom 4]. 
Similarly, if 5 = —6, 
multiplying both sides by 2, we get 
x= —12, [Axiom 3]. 
Again, in the equation 7z—2x%—x=23+15-—10, by collecting 
terms, we have 47 =28. 
an = ls 


Transposition of Terms. 

76. ‘Lo solve 8r—8=2+412. 

This case differs from the preceding in that the unknown 
quantity occurs on both sides of the equation. We can, how- 
ever, transpose any term from one side to the other by simply 
changing its sign. ‘This we proceed to show. 


Subtract « from both sides of the equation, and we get 


38x —x2—8=12, [Axiom 2]. 
Adding 8 to both sides, we have 
32 —x=12+8, [Axiom 1]. 


Thus we see that +z has been removed from one side, and 
appears as —z on the other; and —8 has been removed from 
one side and appears as +8 on the other. 

Hence we may enunciate the following rule: 

Rule. Any term may be transposed from one side of the equation 
to the other by changing its sign. 

It appears from this that we may change the sign of every term 
in an equation ; for this is equivalent to transposing all the terms, 
and then making the right and left hand members change 
places. 


Example. Take the equation —32—-12= w—24. 


Transposing, — £1+24—= 32+12, 
or 3+ 12=—— 7124, 


which is the original equation with the sign of every term changed. 


54 ALGEBRA. (CHAP. 


Tao solves oa 
( eee 4° 

Here it will be convenient to begin by clearing the equation 
of fractiona/ coefficients. This can always be done by multiplying 
both sides of the equation by the least common multiple of the 
denominators. [Axiom 3.] 

Thus, multiplying by 20, 

10% —-60=54+4+ 42 ; 
transposing, 102% —5x—4%=60 ; 
v=60. . 

78, We can now give a general rule for solving any simple 

equation with one unknown quantity. 


Rule, frst, if necessary, clear of fractions; then transpose 
all the terms containing the unknown quantity to one side of the 
equation, and the known quantities to the other. Collect the terms 
on each side, divide both sides by the coefficient of the unknown 
quantity, and the value required vs obtained. 


Hxample 1. Solve 5(a-38)-7(6-—2)+3=24-3(8-2). 
Removing brackets, 52-15 -42+7x+3=24 —24+4 32; 


transposing, 5x+ 7x - 38x =24-24415442-3; 
9x = 54; 
= 0: 


Example 2. Solve (w+1)(2x —1) — 5a = (2% —3)(4% - 5) +47. 
Forming the products, we have 
Qa? +a — 1 — 5a = 2a? -134@+15+447. 
Erasing the term 2z? on each side, and transposing, 
x —5x¢+1384%=15+47+1; 
Sa Cox 
v= 7. 


79, It is extremely useful for the beginner to acquire the 
habit of verifying, that is, proving the truth of his results; 
the habit of applying such tests tends to make the student 
self-reliant and confident in his own accuracy. 

In the case of simple equations we have only to show that 
when we substitute the value of xv in the two sides of the equation 
we obtain the same result. 


| SIMPLE EQUATIONS. 55 


Example. To show that «=7 satisfies the equation 
(2 +1)(2Qe —1) — 5a = (2a —3)(a - 5) +47. 
When 2=7, the left side (w+1)(2~—1)-5a 
=(7+1)(14-—1)-35=(8 x 13) — 35 =69. 
The right side (2% —3)(# —5) +47 
= (14-3)(7 -5) +47 =(11 x 2) +47 =69. 
Thus, since these two results are the same, x=7 satisfies the 
equation. 


EXAMPLES IX, a, 


Write down the solutions of the following equations : 


(Eka & Geers 1b. San 0H = 18: 4.0. °5a= 5. 
Wetec oo = La, 4r=— 19: 8, -—10=—5z2. 
OM az =) 8: Ose 12a = 40ne SL roe — Ov 19 47 50. 
oie = 2b. 1a OS Te. hay thea ie’: 16, 382=— 27. 
ans ete AG. Oars 3, AOS oe lox. 90, —24=- 8x. 
Solve the following equations : 
See Oo tao SD, OO Mor j= 25. 03, iss) 42a: 
DOAN Y= 31a. Obras =i Vs 06. iva ls 2x. 
97, 3x2-18=7-2x. 98. 4e=13-2x-10. 
99, 38¢=7-2x2+8. 80, O=11—-2x2+7- 102. 
81, 8e%-3-5x-5=T7ex. 32, Tx~-13=12-5x- 


88, 5¢—-17+382-5=64-7-8x+115. 

34, 7a—-21-4%+13+4+2e =41 -5x%-7+4+6e. 

85, 15-T7a-9x-28+ 14a -17 =21-3474+13-92%+4+82. 

386. 5x2—-6x2+30-—7x2=2x2+10-7x+ 5x — 20. 

87, 5(v-3)=4(a -2). 38, 11(5- oes — 62). 
09, 3-7(%@-1)=5—4e. ~ 40, 5-4(@-3)=x%-2(a-1). 
4], 8(x-3)-2(8-«a) =2(4+2)-5(5—~2). 

42, 4(5—-—a)-2(@-3)=x2-4-3(4%4+2). 


alee ete ed es al 
43. Pes he 44, 5° fle ote 
2% _l_gi% | pales | ale ear 
45, x aes +7. £40; Mame grt? 
AT. (%+3)(2x —3) — 6x = (x n4)(2x +4) +12. 
48, (x+2 Metal ile 3)(x Qe(a+1)=0. 


= 2) 
49, (2a+1)(2u+6)-7(a-2)=4(a+ Ate 1) ~ Oa, 
50, (8%+1)?+6+18(@+1)? = 9a(8x% -2)+65. 


56 ALGEBRA. [cHAP. 


51, Show that x=5 satisfies the equation 

5a — 6(4 — 4) = 2(4 + 5) + 5(a — 4) -6. 
52, Show that x =15 is the soiution of the equation 

7 (25 — 2) — 2a — 15 = 2(3a — 25) — x. 
53. Verify that x =3 satisfies the equation 

Ha +1)(a+3)+8 = (Qe+1\(a+5). 
54, Show that «=4 satisfies the equation 

(32+ 1)(2% —7) =6(% -3)?+7. 


80. We shall now give some equations of greater difficulty. 


ELxample 1. Solve 5x — (4% —7)(3z — 5) =6 — 3(4x - 9)(a—- 1). 
Simplifying, we have 
5a — (12a? — 41x +35) = 6 -— 3(4x?- 13249); 
and by removing brackets 
7 5x — 1207+ 41x - 35 = 6 — 1247+ 39x — 27. 
Erase the term — 12%* on each side and transpose ; 


thus 5a+41a—-392=6-274+35; 
ioe s 
t= 2, 


Note. Since the — sign before a bracket affects every term within 
it, in the first line of work we do not remove the brackets until we 
have formed the products. 


ies Brad! 
Slee 29 as 
Multiply by 88, the least common multiple of the denominators ; 
352 —- 1l(a-9)=4x”-—44; 
removing brackets, $52-1llz+99=4a%-44; 





Example 2. Solve 4 


transposing, —llx-4x=-44-352-99; 
collecting terms and changing signs, 15a =495 ; 
x = 33. 


Note. In this equation Lad Fe regarded as a single term with 





the minus sign before it. In fact it is equivalent to — (e-9), the 


vinculum or line between the humerator and denominator having the 
same effect as a bracket. [Art. 58.] 


In certain cases it will be found more convenient not to multi- 
ply throughout by the L.c.m. of the denominator, but to clear 
of fractions 1n two or more steps. 


1x.] SIMPLE EQUATIONS. 57 


Example 3. Solve ~= +) 20-3 _5x-32 e+ 9. 
3 35 9 28 
Multiplying throughout by 9, we have 


35 a ey 














transposing, a >in 
5 


Now clear of fractions by multiplying by 5 x 7.x 4 or 140; 
72a — 108 + 45a +405 = 280x — 2800 ; 
2800 — 108 + 405 = 280a — 72a — 45a ; 
3097 = 1632 ; 
Zhe: 


81, Tosolve equations whose coefficients are decimals, we may 
express the decimals as common fractions, and proceed as before; 
but it is often found more simple to work entirely in decimals. 


Example. Solve ‘3752 ~1°875 = '12%+1-185. 


Transposing, *375a - 12% =1'185+41°875 ; 
collecting terms, (°375 — *12)~ =3°06 ; 
that is, 255x = 3°06 ; 
= 3°06 
"255 
= 12, 


EXAMPLES IX, b, 


Solve the equations : 








1, (%+15)(a% - 3) —(x%-3)?=30-15(x—-1). 
Q. 15 -3a =(2x2+1)(2x-1)— (2x —1)(2x2 +3). 
2 Q1—a(2a+1)+2(a—-4)(a +2) =0. 
4, 3(2+5)—3(2a —1)=32—4(x —5)?+ 422, 
5, 3a? -- Ta — (2 +2)(%—2)=(2+1)(%—1) + (a2 -3)(2 +3). 
6, (x-6)(2e-9) - (11 -2x)(7 - x) =5a —4—7(a -2). 
x-1l, «x-9_ CAS - 
Lh Oe tere SEP ger ane 
“+8 5,x-6 62-2 , 3x+5_1 
iT wctetear ek Ook Gish ie 


58 


ALGEBRA. ; [CHAP, IX. 


Solve the equations : 


1a 
13. 
15. 
iy, 
19. 
21. 
22, 
20. 
24. 
25. 
26. 
27, 
28, 
29. 


30, 


dl. 
33, 
30, 
37, 


38. 



































ee oer 2. 12, 2+84+%27=7420. 
2-0 -e=-de>> & Cee al 
yaar ta NU) P(g aa 
+5 ae2+1 _2+3 16 1i-62° 9-72 _5(¢%-1) 
6 9 Wir ; 5 Oot ewe O 
47 — 6x tan ed) 4—52 1-2 -13 
5 en Oe see 18, 6 Fea 
3¢-1 w-1_ 2xa- 31 1 Be SH OF Sa ek 
ees nay PD pees 5) | 4 
“Ww 2 3 hs Satay ver 
3 5 8 5 
P(x = —6) +—. 
$(@-1) ~2(a—4) = 3(e-6) += 
= hs ] 
= =(2ar — -(x-1)-2 
S (w@- 4) ~ 3(2n-9)=3(e -1) 
= (+4) — 5 (w—8) = s(8x ~ 5)-3(« ~ 6) - H(e-2) 
] ] 1 ] 
(3 87} eee Oe 1 
(8 8a) — 51 x) + 5 = 62) 
1 1 Li 1 
2 4 20 — a) =— (5a — 1)-_(5a%- 18) +8. 
Ca )- 7 x) is! ae ) go 3)+8 
e+1 52+9 2+6 x-12 
2 ones ae 
~ l0x+1 x _l3e+4 S(x- 4) 
5 - = 
hn, te 18 4 
sd ae 6 24+5 
a SOS F "i == 1 =e 
ae (2-3) - + 0) + 7 0 
a+4 1] a 1 
ESS e217) a9 M6 Sbay Ss lees 
0 Be a pee Le 
sy eet Li oe Speed | x 
Se = eee ebe hes |. 
na AG a al 5) 
“Ta — 3°35 = 6°4 -— 32a. 33, *Da4+- 254 °141°25= 4a, 
2 20e— lore 94 Ltbx. 34, ‘2x- 0lx+°005z7=11°7. 
Be — 6x = "75x —11. ‘ 96, -4r—'83x2=°7-°3. 
Find the value of x which makes the two expressions 


(8% —-1)(4a-11) and 6(22-1)(@-38) equal. 


What value of « will make the expression 77x — 3(2a — 1)(4a — 2) 
equal to 337 — 8(38%—-1)(@+1)? 


CHAPTER X. 


SYMBOLICAL EXPRESSION. 


62, In solving algebraical problems the chief difficulty of 
the beginner is to express the conditions of the question by 
means of symbols. A question proposed in algebraical symbols 
will frequently be found puzzling, when a similar arithmetical 
question would present no difficulty. Thus, the answer to the 
question “find a number greater than 7 by a” may not be self- 
evident to the beginner, who would of course readily answer an 
analogous arithmetical question, “find a number greater than 50 
by 6.” The process of addition which gives the answer in the 
second case supplies the necessary hint ; and, just as the number 
which is greater than 50 by 6 is 50+6, so the number which is 
greater than 7 by ais xv+a, 


88. The following examples will perhaps be the best intro- 
duction to the subject of this chapter. After the first we leave 
to the student the choice of arithmetical instances, should he 
find them necessary. 


Example 1. By how much does x exceed 17? 


Take a numerical instance ; ‘‘ by how much does 27 exceed 17?” 
The answer obviously is 10, which is equal to 27 ~ 17. 

Hence the excess of x over 17 is x- 17. 

Similarly the defect of x from 17 is 17-2. 


Hxample 2. If x is one part of 45 the other part is 45 — x, 
Hxample 3. If x is one factor of 45 the other factor is 49 
6; 
Example 4. How far can a’man walk in @ hours at the rate of 
4 miles an hour? 
In 1 hour he walks 4 miles, 


In a hours he walks a times as far, that is, 4a miles, 


60 ALGEBRA. [cHAP. 


Example 5. If $20 is divided equally among y persons, the share 


of each is the total sum divided by the number of persons, or ¢ 20. 
y 


Example 6. Out of a purse containing $x and y half-dollars a 
man spends z quarters ; express in cents the sum left. 
$x—4e quarters, 
and y half-dollars=2y quarters; 
.. the sum left=(4”%+2y—z) quarters, 
=25(4%+2y—z) cents. 


EXAMPLES X, a, 


By how much does xz exceed 5? 
By how much is y less than 15? 
What must be added to a to make 7 ? 
What must be added to 6 to make b? 
By what must 5 be multiplied to make a? 
What is the quotient when 3 is divided by a? 
By what must 6x be divided to get 2? 
By how much does 6% exceed 2a ? 
. The sum of two numbers is 2 and one of the numbers is 10; 
what is the other ? 

10, The sum of three numbers is 100; if one of them is 25 and 
another is x, what is the third ? 

11, The product of two factors is 4x; if one of the factors is 4, 
what is the other ? 

12, The product of two numbers is p, and one of them is m3 
what is the other ? 

13, How many times is x contained in 2y ? 

14, The difference of two numbers is 8, and the greater of them 
is @ ; what is the other ? . 

15, The difference of two numbers is x, and the less of them is 6; 
what is the other ? 

16, What number is less than 30 by y? 

17, The sum of 12 equal numbers is 48x; what is the value of 
each number ? 

18, How many numbers each equal to y must be taken to make 
ldxy? 

19, If there are x numbers each equal to 2a, what is their sum ? 

20, If there are 5 numbers each equal to x, what is their product? 


. 


COABDO POOH 


x SYMBOLICAL EXPRESSION. 61 


91, Ifthere are x numbers each equal to p, what is their product ? 


99, If there are n books each worth y dollars, what is the total 
cost ? 


93, If mn books of equal value cost x dollars, what does each 
cost ? 


94, How many books each worth two dollars can be bought 
for y dollars ? 


95, If apples are sold at x for a dime, what will be the cost in 
cents of y apples ? 


96, What is the price in cents of m oranges at six cents a score ? 


97, If I spend n dimes out of a sum of $5, how many dimes 
have I left? 


98, What is the daily wage in dimes of a man who earns $12 
in p weeks, working 6 days a week ? 


99, How many days must a man work in order to earn $6 at 
the rate of y dimes a day ? 


80, If persons combine to pay a bill of $y, what is the share 
of each in dimes ? 


31, How many dimes must a man pay out of a sum of $p so as 
to have left 50x cents ? 


382, How many persons must contribute equally to a fund con- 
sisting of $x, so that the subscription of each may equal y quarters ? 


33. How many hours will it take to travel x miles at 10 miles 
an hour ? 


34, How far can I walk in p~ hours at the rate of g miles an hour ? 

35, If I can walk m miles in n days, what is my rate per day ? 

36. How many days will it take to travel y miles at x miles 
a day? . 

84, We subjoin a few harder examples worked out in full. 


Example 1. What is (1) the sum, (2) the product of three con- 
secutive numbers of which the least is »? 
The two numbers consecutive to n aren+1 andn+2; 
.. the sum=n+(n+4+1)+(n+2) 
=3n+4+3. 
And the product =n(n+1)(n+2). 


Example 2. A boy is x years old, and five years hence his age 
will be half that of his father: how old is the father now? 


In five years the boy will be +5 years old ; therefore his father 
will then be 2(z+5), or 2x+10 years old; his present age must 
therefore be 27+ 10-5 or 2%+5 years. 


62 ALGEBRA. [onar. 


KHxample 3. A and B are playing for money; A begins with $p 
and B with g dimes. B wins $a; express by an equation the fact 
that A has now 3 times as much as B. 


What B has won A has lost ; 
.. A has p—« dollars, that is 10(p—a) dimes, 
B has q dimes + dollars, that is g+ 10x dimes. 
Thus the required equation is 10(p—x) =8(q+10z). 
Hxample 4. A man travels a miles by coach and b miles by train; 


if the coach goes at the rate of 7 miles an hour, and the train at 
the rate of 25 miles per hour, how long does the journey take? 


The coach travels 7 miles in 1 hour ; 


1 
es eae SA errs Be creea 5 hour ; 

2 a 
UAT Ieee ah eee Ch We an: hours. 


Similarly the train travels b miles in = hours. 
5 


a 


“, the whole time occupied is 7+ © hours. 


25 
Example 5. Wow many men will be required to do in p hours 
what g men do in np hours? 
np hours is the time occupied by q men ; 
ne Lk HOUT Aa eas etee eee qx np men; 


thatiis,-7 HOULSs 6), deeeeee ceanes = £2 een. 


P 
Therefore the required number of men is qn. 


EXAMPLES X. b. 


1. Write down three consecutive numbers of which a is the least. 

9, Write down four consecutive numbers of which 0 is the 
greatest. 

3. Write down five consecutive numbers of which c is the 
middle one. 


4, What is the next odd number after 2n-1? 
5, What is the even number next before 2n ? 


6. Write down the product of three odd numbers of which the 
middle one is 2a7+1. 


7, How old is a man who will be x years old in 15 years ? 
8, How old was a man 2 years ago if his present age is m years? 
9, In 2x yearsa man will be y years old, what is his present age ? 


x. ] SYMBOLICAL EXPRESSION. 63 


10, How old is a man who in # years will be twice as old as 
his son now aged 20 years ? 

11. In 5 years a boy will be x years old; what is the present 
age of his father if he is twice as old as his son ? 

12, A has $m and B has n dimes; after A has won 3 dimes 
from B, each has the same amount. Express this in algebraical 
symbols. 

13, A has 25 dollars and B has 138 dollars; after B has won 
x dollars he then has four times as much as A. Express this in 
algebraical symbols. 

14, How many miles can a man walk in 30 minutes if he walks 
1 mile in 2 minutes ? 

15, How many miles can a man walk in 50 minutes if he walks 
x miles in y minutes ? 

16, How long will it take a man to walk p miles if he walks 
15 miles in q hours? 

17, How far can a pigeon fly in x hours at the rate of 2 miles in 
7 minutes ? 

18, A man travels x miles by boat and y miles by train, how 
long will the journey take if the train goes 30 miles and the boat 
10 miles an hour? 

19, If « men do a work in 5z hours, how many men will be 
required to do the same work in y hours ? 

20, How long will it take p men to mow g acres of corn, if each 
man mows 7 acres a day ? 

91, Write down a number which, when divided by a, gives a 
quotient ) and remainder c. 

99. What is the remainder if x divided by y gives a quotient z? 

93, What is the quotient if when m is divided by n there is a 
remainder r? 

94, If a bill is shared equally among n persons, and each pays 
75 cents, how many dollars does the bill amount to ? 

95, A man has $x in his purse, he pays away 25 dimes, and 
receives y cents ; express in dimes the sum he has left. 

96, How many dollars does a man save in a year, if he.earns 
$« a week and spends y quarters a calendar month ? 

97, What is the total cost of 6% nuts and 4% plums, when x 
plums cost a dime and plums are three times as expensive as 
nuts ? bs 

98. If on an average there are x words in a line, and y lines in 
a page, how many pages will be required for a book which contains 
z words ? 


CHAPTER XI. 


PROBLEMS LEADING TO SIMPLE EQUATIONS. 


85, THe principles of the last chapter may now be employed 
to solve various problems. 

The method of procedure is as follows : 

Represent the unknown quantity by a symbol, as 2, and express 
in symbolical language the conditions of the question ; we thus 
obtain a simple equation which can be solved by the methods 
already given in Chapter LX. 


Example I, Find two numbers whose sum is 28, and whose 
difference is 4. 


Let x be the smaller number, then x +4 is the greater. 
Their sum is x+(x%+4), which is to be equal to 28, 


Hence xet+u+4=28; 
Ue = 243 

w= 12, 

and x+4= 16, 


so that the numbers are 12 and 16. 


The beginner is advised to test his solution by finding 
whether it satisfies the conditions of the question or not. 
Example II. Divide $47 between A, B, C, so that A may have 
$10 more than B, and B $8 more than C. 
Let « represent the number of dollars that C has ; then B has 
x+8 dollars,;and A has «+8+10 dollars. 
Hence e+ (#+8) + (#+8+10) =47 ; 
 ¢+%4+84+2748410=47, 
34221 3 
Bs | ; 
so that C has $7, B $15, A $25, 


CHAP, XI.] PROBLEMS LEADING TO SIMPLE EQUATIONS. 65 


EXAMPLES XI, a, 


], Six times a number increased by 11 is equal to 65; find it. 

2, Find a number which when imultiplied by 11 and then 
diminished by 18 is equal to 15. 

3. If 3 be added to a number, and the sum multiplied by 12, the 
result is 84; find the number. 

“ One number exceeds another by 3, and their sum is 27 ; find 

them. 

5, Find two numbers whose sum is 30, and such that one of them 
is greater than the other by 8. 

6, Find two numbers which differ by 10, so that one is three 
times the other. 

7, Find two numbers whose sum is 19, such that one shall exceed 
twice the other by 1. 

8, Find two numbers whose sum shall be 26 and their differ- 
ence 8. 

9, Divide $100 between A and B so that B may have $30 more 
than A. 

10, Divide $66 between A, B, and C so that B may have $8 
more than A, and C $14 more than B. 

11. A, B, and C have $72 among them; C has twice as much 
as B, and B has $4 less than A ; find the share of each. 

12, How must a sum of 78 dollars be divided among A, B, 
and C,so that B may have 8 dollars less than A and 4 dollars 
more than C’? 


Example III, Divide 60 into two parts, so that three times the 
greater may exceed 100 by as much as 8 times the less falls short 
of 200. 

Let x be the greater part, then 60 —-~z is the less. 

Three times the greater part is 3x, and its excess over 100 is 

32 — 100. 
Eight times the less is 8(60- x), and its defect from 200 is 
200 — 8(60 - z). 
Whence the symbolical statement of the question is 
3x — 100 = 200 — 8(60 — x) ; 
3x — 100 = 200 — 480+ 82, 
480 -— 100 — 200 = 8a - 32, 


Da = 180% 
x = 36, the greater part, 
and 60 — ax = 24, the less. 


H.A. E 


66 ALGEBRA. [omar 


Example IV, A is 4 years older than B, and half A’s age 
exceeds one-sixth of 5’s age by 8 years; find their ages, 
Let x be the number of years in B’s age, then A’s age is 2 +4 years. 


One-half of A’s age is represented by $(~+4) years, and one-sixth 
of B’s age by ta years. 


Hence 3(xv+4)-4r=8; 
multiplying by 6 3x+12-—x%=48; 
Vi oe 

hoped Noe 


Thus B’s age is 18 years, and A’s age is 22 years. 


13, Divide 75 into two parts, so that three times one part may 
be double of the other. 

14, Divide 122 into two parts, such that one may be as much 
above 72 as twice the other is below 60. 

15, A certain number is doubled and then increased by 5, and 
the result is less by 1 than three times the number ; find it. 

16, How much must be added to 28 so that the resulting number 
may be 8 times the added part ? 

17, Find the number whose double exceeds its half by 9. 

18, What is the number whose seventh part exceeds its eighth 
part by 1? 

19, Divide 48 into two parts, so that one part may be three-fifths 
of the other. 

90. If A, B, and C have $76 between them, and A’s money is 
double of B’s and C’s one-sixth of B’s, what is the share of each? 

91, Divide $511 between A, B, and C, so that B’s share shall be 
one-third of A’s, and C’s share three-fourths of A’s and B’s together. 

29, B is 16 years younger than A, and one-half 5’s age is equal 
to one-third of A’s ; how old are they ? 

93, A is 8 years younger than J, and 24 years older than C; 
one-sixth of A’s age, one-half of 5’s, and one-third of C’s together 
amount to 38 years ; find their ages. 

94, Find two consecutive numbers whose product exceeds the 
square of the smaller by 7. [See Art. 84, Ex. 1.] 

95, The difference between the squares of two consecutive 
numbers is 31 ; find the numbers. 


86. We shall now give examples of somewhat greater 
difficulty. 
Example I. A has $6, and B has six dimes; after B has won 


from A acertain sum, A has then five-sixths of what B has; how 
much did B win ? 


XI.]} PROBLEMS LEADING TO SIMPLE EQUATIONS. 67 


Suppose that B wins # dimes, A has then 60—«a dimes, and B 
has 6+” dimes. 


Hence 60—»=2(6+4+2); 
360-—6x—= 30+ 52, 
Lins sa0% 
7=380. 


Therefore B wins 30 dimes, or $3. 


Example II. A is twice as old as B, ten years ago he was four 
times as old ; what are their present ages ? 
Let B’s age be x years, then A’s age is 2% years. 
Ten years ago their ages were respectively «—10 and 27%—10 
years ; thus we have 
2x—10=4(x%—10) ; 
2x—10=4a—40, 
2x%—= 30 ; 
a 15, 


- so that Bis 15 years old, A 380 years. 


EXAMPLES XI. b. 


1, A has $12 and Bhas $8; after B has lost a certain sum to A 
his money is only three-sevenths of A’s; how much did A win ? 

9, Aand B begin to play each with $15; if they play till B’s 
money is four-elevenths of A’s, what does B lose ? 

8, Aand B have $28 between them; A gives $3 to B and then 
finds he has six times as much money as B; how much had each 
at first ? 

4, Ahad three times as much money as B; after giving $3 to 
B he had only twice as much; what had each at first ? 

5, A father is four times as old as his son; in 16 years he will 
only be twice as old ; find their ages. 

6. A is 20 years older than B, and 5 years ago A was twice as 
old as B; find their ages. 

7, How old is a man whose age 10 years ago was three-eighths 
of what it will be in 15 years ? 

8, Ais twice as old as‘’B; 5 years ago he was three times as 
old ; what are their present ages ? 

9, A father is 24 years older than his son; in 7 years the son’s 
age will be two-fifths of his father’s age; what are their present 
ages ? 


68 ALGEBRA. [cuap. 


Example III. A person spent $56.40 in buying geese and 
ducks ; if each goose cost 7 dimes, and each duck 3 dimes, and if 
the total number of birds bought was 108, how many of each did 
he buy ? 

In questions of this kind it is of essential importance to have all 
quantities expressed in the same denomination; in the present 
instance it will be convenient to express the money in dimes. 

Let « be the number of geese, then 108—~ is the number of ducks, 

Since each goose costs 7 dimes, x geese cost 7x dimes. 

And since each duck costs 3 dimes, 108—w ducks cost 8(108—~) 
dimes, 

Therefore the amount spent is 


7x+8(108—«a) dimes. 


But the question states that the amount is also $56.40, that is 564 
dimes. 


Hence 72+8(108—2) =564 ; 
74+ 824 —3x=564, 
4%=240, 

*, =60, the number of geese, 

and 108—x=48, the number of ducks. 


Note. In all these examples it should be noticed that the un- 
known quantity x represents a number of dollars, ducks, years, etc. ; 
and the student must be careful to avoid beginning a solution with 
a supposition of the kind, ‘‘let x=A’s share”’’ or ‘‘let x=the 
ducks,’’ or any statement so vague and inexact. 


It will sometimes be found easier not to put x equal to the 
quantity directly required, but to some other quantity involved 
in the question; by this means the equation is often simplified. 


Example IV. A woman spends $1 in buying eggs, and finds 
that 9 of them cost as much over 25 cents as 16 cost under 75 cents ; 
how many eggs did she buy ? 

Let x be the price of an egg in cents; then 9 eggs cost 9% cents, 
and 16 eggs cost 16x cents ; 


oe 9x%—25=—75—162, 
’ 25x%= 100 5 
eA 
Thus the price of an egg is 4 cents, and the number of eggs 
=100+4=25. 
10. A sum of $30 is divided between 50 men and women, the 


men each receiving 75 cents, and the women 50 cents; find the 
number of each sex. 


XI.] PROBLEMS LEADING TO SIMPLE EQUATIONS... 69 


11, The price of 15 yards of cloth is as much less than $10 as 
the price of 27 yards exceeds $20; find the price per yard. 


12, A hundredweight of tea, worth $68, is made up of two 
sorts, part worth 80 cents a pound and the rest worth 50 cents a 
pound ; how much is there of each sort ? 


138, A man is hired for 60 days on condition that for each day 
he works he shall receive $2, but for each day that he is idle he 
shall pay $1 for his board: at the end he received $90; how many 
days had he worked ? 


14, A sum of $6 is made up of 50 coins, which are either quar- 
ters or dimes ; how many are there of each ? 


15, A sum of $11.45 was paid in half-dollars, quarters, and 
dimes ; the number of half-dollars used was four times the number 
of quarters and ten times the number of dimes; how many were 
there of each ? 


16, A person buys coffee and tea at 40 cents and 80 cents a 
pound respectively ; he spends $15.10, and in all gets 24 lbs. ; how 
much of each did he buy ? 


17, A man sold a horse fora sum of money which was greater 
by $68 than half the price he paid for it, and gained thereby $18; 
what did he pay for the horse ? 


18, Two boys have 240 marbles between them; one arranges 
his in heaps of 6 each, the other in heaps of 9 each. There are 36 
heaps altogether ; how many marbles has each ? 


19, A man’s age is four times the combined ages of his two sons, 
one of whom is three times as old as the other; in 24 years their 
combined ages will be 12 years less than their father’s age ; find 
their respective ages. 


990, Asum of money is divided between three persons, A, B, and 
C, in such a way that A and B have $42 between them, B and C 
have $45, and C and A have $53; what is the share of each ? 


91, A person bought a number of oranges for $3, and finds that 
12 of them cost as much over 24 cents as 16 of them cost under 60 
cents ; how many oranges were bought ? 


99, By buying eggs at 15 for a quarter and selling them at a 
dozen for 15 cents a man lost $1.50; find the number of eggs. 


93, I bought a certain number of apples at four for a cent, and 
three-fifths of that number at three for acent; by selling them at 
sixteen for five cents I gained 4 cents ; how many apples did I buy ? 


94, If 8 lbs. of tea and 24 lbs. of sugar cost $7.20, and if 3 lbs. 
of tea cost as much as 45 lbs. of sugar, find the price of each per 
pound. 


70 ALGEBRA. [CHAP. XI. 


95, Four dozen of port and three dozen of sherry cost $89 ; if 
a bottle of port costs 25 cents more than a bottle of sherry, find the 
price of each per dozen. 


96, Aman sells 50 acres more than the fourth part of his farm 
and has remaining 10 acres less than the third; find the number 
of acres in the farm. 


97, Find a number such that if we divide it by 10 and then 
divide 10 by the number and add the quotients, we obtain a result 
which is equal to the quotient of the number increased by 20 when 
divided by 10. 


98, A sum of money is divided between three persons, A, B, 
and CO, in such a way that A receives $10 more than one-half of 
the entire amount, B receives $10 more than one-third, and C the 
remainder, which is $10; find the amounts received by A and B. 


99, The difference between two numbers is 15, and the quotient 
arising from dividing the greater by the less is 4; find the numbers. 


380, A person in buying silk found that if he should pay $3.50 
per yard he would lack $15 of having money enough to pay for it ; 
he therefore purchased an inferior quality at $2.50 per yard and 
had $25 left ; how many yards did he buy ? 


381, Find two numbers which are to each other as 2 to 3, and 
whose sum is 100. 


382, A man’s age is twice the combined ages of his three sons, 
the eldest of whom is 3 times as old as the youngest and 3 times as 
old as the second son; in 10 years their combined ages will be 4 
years less than their father’s age ; find their respective ages. 


338, The sum of $34.50 was given to some men, women, and 
children, each man receiving $2, each woman $1, and each child 
50 eents. The number of men was 4 less than twice the number of 
women, and the number of children was 1 more than twice the 
number of women ; find the total number of persons. 


34, A man bought a number of apples at the rate of 5 for 
3 cents. He sold four-fifths of them at 4 for 8 cents and the 
remainder at 2 for a cent, gaining 10 cents; how many did he 
buy ? 

35, <A farm of 350 acres was owned by four men, 4, B, C, and 
D. B owns five-sixths as much as A, C four-fifths as much as B, 
and D one-sixth as much as A, B, and C together; find the num- 
ber of acres owned by each. 


CHAPTER XIL 


ELEMENTARY FRACTIONS. 


Highest Common Factor of Simple Expressions, 


68. Derrinition. The highest common factor of two or 
more algebraical expressions is the expression of highest dimen- 
sions [Art. 68] which divides each of them without remainder. 

The abbreviation H.C.F. is sometimes used instead of the 
words highest common factor. 


89, In the case of simple expressions the highest common 
factor can be written down by inspection. 


2 
oO 


Example 1, The highest common factor of a4, a?, a7, a° is a. 
9 2 > 


Example 2. The highest common factor of a?b4, a*b°c?, atb’c is 
a*b*; for a? is the highest power of a that will divide a’, a?, at; 
b+ is the highest power of b that will divide b+, b°, b7; and c is not a 
common factor. 


90, If the expressions have numerical coefficients, find by 
Arithmetic their greatest common measure, and prefix it as a 
coefficient to the algebraical highest common factor. 


Hxample. The highest common factor of 2latz*y, 35a7xty, 28a%ay 
is Ja*xy ; for it consists of the product of 
(1) the greatest common measure of the numerical coefficients ; 


(2) the highest power of each letter which divides every one of 
the given expressions. 


EXAMPLES XII. a. 


Find the highest common factor of 


eeoceD aur, Doc Ary, ,. 0c’, 50°C. PT ns hea Verne 
Dewar cn auc.) Ga.eaco,Oabe. {, 6a*ytz, Qay. 8, lby®, Sry fe, 
9, 12a%bc?, 18ab?c, Oe iecyr a: Zl aae. 

1]. 8ax, 6a*y, 10ab?x?, Lee oes OF cary”, 

18, 14bc?, 63ba?, 56b7c. 14, lary, 60a5y72*, 25022", 


15, Ixy’, Slayz, 34a°yz «16, Tia°bic?, B3a%%c%, ab ct, 


72 ALGEBRA. [CHAP. 


Lowest Common Multiple of Simple Expressions, 


91, Derrryirion. The lowest common multiple of two or 
more algebraical expressions is the expression of lowest dimen- 
sions which is divisible by each of them without remainder. 

The abbreviation L.C.M. is sometimes used instead of the 
words lowest common multiple. 


92. In the case of simple expressions the lowest common 
multiple can be written down by inspection. 
Hxample 1. The lowest common multiple of a4, a3, a?, a® is a®, 


Example 2. The lowest common multiple of ab‘, ab®, a?b7 is 
a®b’; for a? is the lowest power of a that is divisible by each of the 
quantities a®, a, a7; and b7 is the lowest power of b that is divisible 
by each of the quantities b*, b°, b’. 


93, If the expressions have numerical coefficients, find by 
Arithmetic their least common multiple, and prefix it as a co- 
efficient to the algebraical lowest common multiple. 


Example. The lowest common multiple of 2latz*y, 35a°aty, 
28a°®xy is 420ata4y ; for it consists of the product of 


(1) the least common multiple of the numerical coefficients ; 


(2) the lowest power of each letter which is divisible by every 
power of that letter occurring in the given expressions. 


EXAMPLES XII. b, 


Find the lowest common multiple of 


Reo, oY, 2, abt abe, Bee ory er. 

A 4a, Baber. 5, 4a4bc?, 5ab?. 6. 2ab, 4ay. 

7, mn, ni, lm. OP yy oor eee: 9, xy, 8yz, 4zx. 
LOmee oT Ag pt Pg: ell ploa ys barye. 1357 9ab?, 2107e: 
13. 27a?; 816%, 18076": [Ae har Gey ait 6 
Lo mewod 0, 2007 yo0z. 16. 72p79°r4, 108p%97r. 


Find both the highest common factor and the lowest common 
multiple of 


17, 2ab?, 3a2b3, 4a4b. 18, 1523 y?, 5a?yz°. 19, 2a4, 8a2h3c7, 
OD eS ia coe. 21, 32a*b®c, 48a7bc°. 
99, 5lim®p?, pn, 34mnp'. 93, 49a4, 56b4c, 2lac’., 


94, 66a°*b%cat, 55ab’ayrz, 12a? yz’. 


XII] ELEMENTARY FRACTIONS. 73 


Elementary Fractions, 


94, Derrixition. If a quantity w be divided into b equal 
parts, and a@ of these parts be taken, the result is called the 


Jraction ; of x. If « be the unit, the fraction of w is called 


simply “the fraction @ ”+. go that the fraction 
ply b 


b 


represents a 


equal parts, b of which make up the unit. 


95, In this chapter we propose to deal only with the easier 
kinds of fractions, where the numerator and denominator are 
simple expressions. 

Their reduction and simplification will be performed by the 
usual arithmetical rules. For the proofs of these rules the 


reader is referred to the Elementary Algebra for WSchools, 
Chapter xv. : 


Rule, Zo reduce a fraction to rts lowest terms: divide 
numerator and denominator by every factor which is common to 
them both, that 1s by their highest common factor. 

Dividing numerator and denominator of a fraction by a com- 
mon factor is called cancelling that factor. 


ESOS ae Lee 


35a°b%c _ 5atb _ 


4 
Tab-c 1 oe 





EXAMPLES XII, c, 


Reduce to lowest terms : 


























2.3 oe 3 

2203 3/2 af 232 
d. “Tea ; 10. eye H. yee La. iba 
a Beaty Eee ag Ses ag Be 


74 ALGEBRA. [CHAP. 


Multiplication and Division of Fractions. 


96. Rule, Zo multiply algebraical fractions: as in Arith- 
metic, multiply together all the numerators for a new numerator, 
and all the denominators for a new denonunator. 


Hxample | 2a x 5a" v, BD? (QO Dae oUe ae 
é . 3b 2a7b on ~ 3b x 902 x on oq,” 
by cancelling like factors in numerator and denominator. 


8a°b a 


Do Oras 0. 
all the factors cancelling each other. 


97, Rule, Zo divide one fraction by another: invert the 
divisor and proceed as in multiplication. 

Ta? Boke, 28a%e? 

4a3y? 5ab?  15b%xry? 
= Ta? x Beta ——— 15b*xy? 

42°y2" Bab? 28a? 





Hxample 2. 











Example. 


all the other factors cancelling each other, 


EXAMPLES XII. d. 


Simplify the following expressions : 



































ay, a°b? ab . 4c*d FLEE lie ANE 
1. ab ‘ye 2. Qed abe 3: 3y°z * date 
Gara? 140% 3ab? 15b7c* Tee eee 
= ; SK pees fic ® : 
e Tab? c l2ax 5. 5b’c— 9a2b* 6, Bbc? ‘ l4be 
7 am ,, 2cd?  Dmy, g 4a*b , 3p7q? . PY 
' By *3ab 4am? Yay * Babb? a2 ee 
9 20% ,, 100? «te i 10 Uae iW Oe 34/3 ye 
' Bat de®” 308" "gad ged” oe by 
Sore") Sax? xy? 15b2_ 14d? , 81d? 
x . cr ° 
Ll. aby * Bate * by 12. Fc * abe BIA 


Reduction to a Common Denominator, 


‘98, In order to find the sum or difference of any fractions, 
we must, as in Arithmetic, first reduce them to a common 
denominator ; and it is most convenient to take the lowest com- 
mon multiple of the denominators of the given fractions. 


XI. J ELEMENTARY FRACTIONS. 75 


Hxample. Express with lowest common denominator the fractions 
age anak 
Say’ 6xyz Qyz 
The lowest common multiple of the denominators is 6ayz. Multi- 
plying the numerator of each fraction by the factor which is required 
to make its denominator 6xyz, we have the equivalent fractions 
20 eee 3cx 
6xyz 6xyz Bayz 
Note. The same result would clearly be obtained by dividing the 
lowest common denominator by each of the denominators in turn, 
and multiplying the corresponding numerators by the respective 
quotients. 





EXAMPLES XII, e. 


Express as equivalent fractions with common denominator : 


Tee oer Be a 4, = 
Bay ae 6. ay He lh @ = 8, a = 
9. ? & 105); Qa. Lge ae - 
Bowe Bowe wWoaet 


Addition and Subtraction of Fractions. 


99, Rule. To add or subtract fractions: express all the 
fractions with their lowest common denominator ; form the algebrat- 
cal sum of the numerators, and retain the common denominator. 

5x 712 


. Simplify 27 4?x-/™, 
Example 1. Simplify 5 sie 7 


The least common denominator is 12. 
202+92—l4a_15x”_ 5x 











The expression = 12 =i puepie 
» ae 3a ab ab 
Example 2. Simplify Pe Fy 


76 ALGEBRA. [CHAP. XII. 


ey 
azc2 3ca® 


. axe — C2 : Wek : 
The expression = ee and admits of no further simplification. 
CG: Ce 


Example 3. Simplify 


Note. The beginner must be careful to distinguish between 
erasing equal terms with different signs, as in Example 2, and 
cancelling equal factors in the course of multiplication, or in 
reducing fractions to lowest terms. Moreover, in simplifying frac- 
tions he must remember that a factor can only be removed from 
numerator and denominator when it divides each taken as a whole. 
6ax —cy 
cae 
and not the whole numerator. Similarly a cannot be cancelled 
because it only divides 6ax% and not the whole numerator. The 
fraction is therefore in its simplest form. 


Thus in c cannot be cancelled because it only divides cy 


When no denominator is expressed the denominator 1 may 
be understood. 
te OL Ooty a0 
Example 4. ee ee ee Be 
xcample 3x dy elias vy, 
If a fraction is not in its lowest terms it should be simplified 
before combining it with other fractions. 


vi : at ay _ax x _sav—2Qe 
xample 5 5 aay a 5 


EXAMPLES XII. f. 


Simplify the following expressions : 








ee ee ee oe 
5 5-5 DD Gey TB rap Bh Bom 
9, ate 10. os 11, aoe 12, re 
Ry Sa ee 14, Pees 15, 18-2 
Be ee uae ae 
19.6 2 20 en Bes Sheen 
93, oe dx, xy OA, a= a? ae 





~ dy t By 3a ab? 6be 


CHAPTER] XLT, 


SIMULTANEOUS EQUATIONS, 


100, ConstpEr the equation 22+ 5y=28, which contains two 
unknown quantities. 


By transposition we get 
By =23 — 2x ; 


that is, a= abs nd, RR (1). 





From this it appears that for every value we choose to give 
to # there will be one corresponding value of y. Thus we shall 
be able to find as many pairs of values as we please which 
satisfy the given equation, 


For instance, if v=1, then from (1) we obtain 2 
Again, if = —2, then ya ; and so on. 


But if also we have a second equation containing the same un- 
known quantities, such as 8x+4y=24, 





we have from this y 7 Ree (2). 


If now we seek values of w and y which satisfy both equa- 
tions, the values of y in (1) and (2) must be identical. 





Therefore pe ee era, 
5 4 
Multiplying across 92-—8r=120—-152; 
(LZ 5 
xv=A4, 
Substituting this value in the first equation, we have 
8 + 5y = 23 ; 
w. SY=15; 
con ah 
and c=4, 


Thus, if both equations are to be satisfied by the same values 
of # and y, there is only one solution possible. 


78 ALGEBRA. [CHAP. 


101, Derrryition. When two or more equations are satisfied 
by the same values of the unknown quantities they are called 
simultaneous equations, 

We proceed to explain the different methods for solving simul- 
taneous equations. In the present chapter we shall contine our 
attention to the simpler cases In which the unknown quantities 
are involved in the first degree. 


102, In the example already worked we have used the 
method of solution which best illustrates the meaning of the 
term svmultaneous equation ; but in practice it will be found that 
this is rarely the readiest. mode of solution. It must be borne 
in mind that since the two equations are simultaneously true, 
any equation formed by combining them will be satisfied by the 
values of x and y which satisfy the original equations. Our 
object will always be to obtain an equation which involves one 
only of the unknown quantities. 


1038. The process by which we cause either of the un- 
known quantities to disappear is called elimination. We shall 
consider two methods. 


Elimination by Addition or Subtraction. 


Example 1. Solve BOA TY =O) senna ote shee eee (1); 
BiH 2y = 16.20 oes ue oe ee ee ee (2). 
To eliminate x we multiply (1) by 5 and (2) by 8, so as to make 
the coefficients of 2 in both equations equal. This gives 
15x + 35y = 135, 


. ldx+ by = 48 ; 
subtracting, 297 = 877; 
i MN oa; 


To find x, substitute this value of y in ether of the given 
equations. 


Thus from (1), Noe Spd Ri Ear 
meas 
and. y = tf 
Note. When one of the unknowns has been found, it is immaterial 


which of the equations we use to complete the solution. Thus, in 
the present example, if we substitute 3 for y in (2), we have 


5a+6=16; 
a = 2, as before. 


XIII. ] SIMULTANEOUS EQUATIONS. 79 


Example 2. Solve PRS ick LIE ech ESE ee eee (1), 
aoe ELEM Ae agile pe cor 


Here it will be more convenient to eliminate y. 
Multiplying (1) by 2, l4x+4y = 94, 


and from (2) 5a —4y=13 
adding, 19% = 95 ; 
Goa 
Substitute this value in (1), 
30 + 2y = 47 ; 
. y=6, 


Note. Add when the coefficients of one unknown are equal and 
unlike in sign ; subtract when the coefficients are equal and like in 
sign. 

Elimination by Substitution. 


Example 3. Solve Lat NO ah cise aeantiaate ks soe nite a asst ie Ag 
QE OY aieniians wie NE he cote en stens (2). 


Here we can eliminate x by substituting in (2) its value obtained 
from (1). Thus 


24 — T5y +1) = By; 
48 — 35y —7 = by ;5x 
4] =41y; 
noe fit 
and from (1) La. 


104, Any one of the methods given above will be found 
sufficient ; but there are certain arithmetical artifices which 
will rnorincn shorten the work. 


Hxample. Solve DSA cO UY eat anon Sar Enki eigs ei tst ete reeset ‘ae 


Noticing that 28 and 63 contain a common factor 7, we shall make 
the coefficients of x in the two equations equal to the least common 
multiple of 28 and 63 if we multiply (1) by 9 and (2) by 4. 


Thus 252ax —207y = 198, 

252% —-220y = 68; 

subtracting, 13y = 130; 
that is, y = 10, 


and therefore from (1), x= 9, 


[CHAP. 


80 ALGEBRA. 


EXAMPLES XIII. a. 


Solve the equations: 


l, x«+y=19, 9. ety = 23, oO ety=ail; 
See TE Li Yeo. ey = —9, 
4, x+y = 24, he, ey, G6, w—y= 25, 
x-y= 0. et 8/0. cry = 13, 
7, 3a2+5y = 50, 8, x+5y= 18, 9, 4%+ y=10, 
44+ 3y = 41. 3a+2y = 41. 5u+Ty = 47. 
10, 7 -6y = 25, ll, 5e+4y=7, 12, 3¢—Ty=1 
5a+4y = 51. 4x + 5y = 2. 4x+ y= 53. 
18, 7x+5y = 45, 14, 4¢+5y =4, Lijelig= fy at433 
Qu - 3y = 4. 5x — 3y = 79. 2x — 3y = 13. 
16, 4x-3y=0, 17, -2x+3y = 22, 18, 72+3y = 65, 
7a —4y = 36. 5a +2y = 0. iz —8y = 32. 
19, 1l3e-y=14, 20. 9ux-8y = 14, Q1, 14%+13y = 35, 
2a —7Ty = 9. 15x - 14y = 20. 21x+19y = 56. 
0D. bf ay 21, Doaoot = sor, 04, bdea7y= it, 
21x —9y = 75. 10x” = Ty — 15. etn a bap 
95, 138x-9y=46, 26, 6x-5y=11, 27, lly-llx=66, 
llz-12y=17. 28a+21y = 7. 7x +8y = 3. 
98, 6y-5x =11, 29, 32+10=5y, 80, 4y =47+4+32, 
4a = Ty — 22. Ty = 4¢+13. 5a = 30- ldy. 
ol, 12+138y 57, Cols lly ieee Glee 
13x4+ lly = 17, 292 — 39y = 17. 4lz+37y = 17. 
105, We add a few cases in which, before proceeding to 
solve, it will be necessary to simplify the equations. 
Example, Solve 65(%+2y)—(38a+ ly) =14 crrcceccccceccceneeees (1); 
Ja Oy Sie — dy) a 38 ase ee (2), 
From (1), 52+10y - 3x -lly=14; 
Dae aap ct We ee vargas eaenecne toes (3). 
From (2), 7x —9y -3x+12y = 38; 
Ape 38 Ure, ee (4). 
From (3), 6a — 3y = 42, 


By addition, 10z = 80; whence x = 8, 


From (8) we obtain y = 2. 


xu. ] SIMULTANEOUS EQUATIONS. 81 


106, Sometimes the value of the second unknown is more 
easily found by elimination than by substituting the value of 
the unknown already found. 











- Example. Solve 3a —4 ie = =e Lek hy a eee Cy 
i 
By +4 — lig. _ aie a 
5 3(2e Weare Ieee cartes veer ecnecises es Cay 
Clear of fractions. Thus 
from (1), 42a —2y +10 = 287-21; 
VA Ga Le NO has cnt casiceaee ate arse (3). 
From (2), 9y +12 --10%+25 = ldy ; 
LOS UG Spencer yes clea castes steeds ev (4). 
Eliminating y from (38) and (4), we find that 
oe 
13 
Eliminating x from (3) and (4), we find that 
y 2 20! 
26 


107. Simultaneous equations may often be conveniently 


ee 1 uh ee 
solved by considering — and ~ as the unknown quantities. 
we o 


Example. Solve 2 se Sat LO come nat teestie Sin aennces Rese tee ce ‘id Ws 
Ae ey 
EOE TER else en: Ne ee OP (2) 
te Ms 
Multiply (1) by 2 and (2) by 3; thus 
16_18_y 
oY 
30 | 18 = 9A > 
og wes 
adding, a0 8 ; 
% 
multiplying up, 46 = 232 ; 
fig A 


and by substituting in(1), 4 =3. 
HLA. F 


82 ALGEBRA. [cuar. 
EXAMPLES XIII. b. 

Solve the equations : 

1, 2x-y=4, 9, 4e-y=1, 3. £+2y = 13, 
Bed Ep eed ee US 
Did en ne 35 
HA a eye Bx a 

4, eae 5, 2 =O, 6, eae G 
%+3y = 2 4x -3y=1 4x + 5y = 0. 

Ty bare Ay, 8, «x-y=9, Q, x+y=-2, 
4x By _ Bd 08 e1Y_o 
i ae ye any 46 
1 3 ] ] 
Ee =I, é Sop oD) = 20); ‘ Rig 20) = 

10; 5(v +3) 0 11 pe ~ 2y 0 12 ate al 0 
sey = 4h s(y-+8) = 2. Sm = Dy 

13. 3(v@-y)+2%a+y)=15, 38(at+y)+2(x-y) = 25. 

14, 3(@+y-5)=22(y—-x), 3(v-—y-7)+2(r4+y-— 2) =0. 

15, 4(2x-y-6)=38(8%-2y-5), 2x-y+1)+4x = 3y+4. 

16. 7(2a—y)+5(8y—4x7)+30=0, 5(y-—2+3) = 6(y— 22). 
+4 y-4_ oe le ya lis ot oy 

17. yma ee 18. 25 

peak 90, 24° =39, Ole eae 
cy “x sy eae ats 
bigs 708 Big SoEE 
aay ay “x y 














108, In order to solve simultaneous equations which contain 
two unknown quantities we have seen that we must have two 


equations. 


Rule. 


be solved by the refes already given. 


Similarly we find that in .erder to solve simul- 
taneous equations which contain three unknown quantities we 
must have three equations. 


Eliminate one of the unknowns from any pair of the 
equations, and then eliminate the same unknown from another pair. 
Two equations involving two unknowns are thus obtained, which may 


The remaining unknown is 


then found by substituting in any one of the given equations. 


XIII. | SIMULTANEOUS EQUATIONS. 83 


Example. Solve PRO Orees etme coe Pee bess saa seecdaceeerts (1), 
Miri as Vice go ycame 1 uae ae fo ane eee Retry (2), 
DOE EA Ber SN es ys as bog teeceaiy orsniehaanes (8). 


Choose y as the unknown to be eliminated. 
Multiply (2) by 5, 20%+10y-15z=0; 
Multiply (1) by 2, 14%+10y-—142= - 16; 


by subtraction, O75 —@ 1 On see testare sees keoe' beuvce, (E)e 
Multiply (2) by 2, 8x+4y-62=0; 

from (3), 5a — 4y + 42 = 35 5 

by addition, 13x — 2z = 35. 
Multiply (4) by 2, 122% —2z = 32; 

by subtraction, Sr Ds 
From (4) we find a= 2, 

and from (2), y=-3. 


109. Some modification of the foregoing rule may often be 
used with advantage. 


Example. Sol ees ba tO. 
xample olve 5 re id 
ja teen as 
379 2 


From the equation 5 Sle a+ 1, 


we have a 2) deride eral oA ee Peccceeeseae eomeeh (1). 
Also from the equation 57 is +2, 
we have rh Mio) ah re aE Nec tence an odds Aras avosalloas ieee (2). 


And from the equation at5e 13, 
we have Ri aa eae) ee ee Tatar ceestbycaetes tas (3). 


Eliminating z from (2) and (3), we have 
21x +4y = 282 ; 
and from (1} 12x -4y = 48 ; 
whence #=10, y=18. Also by substitution in (2) we obtain z= 14, 


84 


ALGEBRA. 


EXAMPLES XIII. 


Solve the equations : 


Ub, 


co 


Lt, 
12, 
13, 
14. 
15, 
16, 


3x —2y+2= 4, Dh. 
2x + 3y —2= 3, 

Lt Y+tZa S, 

xv + 2y +32 = 32, 4, 


da — dy + 6z.=.27, 

7x +8y —9z = 14, 

7«—4y-3z= 0, 6, 
5x — 3y + 22 = 12, 

34+2y-5z= 0. 

3y — 62 —5a = 4, 8, 
22 -3dx- y=8, 

x —2y+22+2=0. 


10, 


dar (zy) = 11, 


Qe sey) =2, Haty) = 78-2), 





: 
3 


(CHAP. XIII. 


ox + 4y — 6z = 16, 
4eu+ y-— z= 24, 
x —d3y—225 

LNs Rp Sema 4¥, 
62 + dy + 2z = 84, 
da + 4y — 5z = 13. 
4x+3y- z= 9, 
9G = Omer} 
wAYy — 32 =) 2. 
by +22 +52 = 21, 
82 — 32 9 = 8, 
22 +22 -3y = 39. 


(z-4x) =y. 


w= 4y +32. 


GCHAPTER: XLV; 
PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS. 


1190, Ivy the Examples discussed in the last chapter we have 
seen that it is essential to have as many equations as there are 
unknown quantities to determine. Consequently the statement 
of problems which give rise to simultaneous equations must 
contain as many Independent conditions, or different relations 
between the unknown quantities, as there are quantities to be 
determined. 


Example 1. Find two numbers whose difference is 11, and one- 
fifth of whose sum is 9. 


Let x be the greater number, y the less ; 





then yt en atta saree a ae eres § Gh): 
Also oe =o, 
or GY = 40 S232 head eee toe (2). 


By addition 22=56 ; and by subtraction 2y=34. 
The numbers are therefore 28 and 17. 


Example 2. If 15 lbs. of tea and 10 lbs. of coffee together cost 
$15.50, and 25 lbs. of tea and 13 lbs. of coffee together cost $24.55 ; 
find the price of each per pound. 


Suppose a pound. of tea to cost « cents, 


a eet COMLCE aie iae Ys aaa ele 
Then from the question we have 
1g Oy = la bW aes aes weNierds ables bia 8 Ly, 
PAVERS GD ERO 115 Diya eer ey ee (2). 


Multiplying (1) by 5 and (2) by 38, we have 
752+ 50y=7750, 
752+ 89y =7365. 


Subtracting, weLiv— ooo. 
Y= 35. 
And from (1), 15% +850 = 1550 ; 
whence tba 1200 5 
MzaS0): 


.*. the cost of a pound of teais 80 cents, . 
and the cost of a pound of coffee is 35 cents. 


86 ALGEBRA. [CHAP. 


Hxample 3. In a bag containing black and white balls, half the 
number of white is equal to a third of the number of black ; and 
twice the whole number of balls exceeds three times the number of 
black balls by four. How many balls did the bag contain ? 


Let x be the number of white balls, and y the number of black — 
balls; then the bag contains 7 +y balls. 


We have the following equations : 


pag AI HO Neh RAM, l 
2 3 es. ( ) 
2A Y) =H OY EE vacenceatessesessescescststeaae (2). 


Substituting from (1) in 2, we obtain 
AY 4 y= By +4; 
whence Uai2: 


and from {1), x= 8. 
Thus there are 8 white and 12 black balls. 


111, In a problem involving the digits of a number the 
student should carefully notice the way in which the value of a 
number is algebraically expressed in terms of its digits. 


Consider a number of three digits such as 435; its value is 
4x100+3x10+5. Similarly a number whose digits beginning 
from the left are x, y, z 

=x hundreds+y tens+z units 


= 100% + 10y +2. 


Hxample. A certain number of two digits is three times the sum 
of its digits, and if 45 be added to it the digits will be reversed ; 
tind the number. 


Let x be the digit in the tens’ place, y the digit in the units’ place ; 
then the number will be represented by 10x+y, and the number 
formed by reversing the digits will be represented by 10y+2. 


Hence we have the two equations 


LODE 83 (20) ee coetans cout oe eee (1), 
_and 10% dD = LOG 12s as ssi vacks ees creek wares (2). 
From (1), Ley 
from (2), yee), 


From these equations we obtain x = 2, y = 7. 
Thus the number is 27. 


XIv.] PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS. 87 


EXAMPLES XIV. 


1, Find two numbers whose sum is 54, and whose difference is 
12 


foal 


9, Thesum of two numbers is 97 and their difference is 51; find 
the numbers. 


8. One-fifth of the difference of two numbers is 3, and one-third 
of their sum is 17; find the numbers. 


4, One-sixth of the sum of two numbers is 14, and half their 
difference is 13; find the numbers. 


5, ‘Four sheep and seven cows are worth $131, while three cows 
and five sheep are worth $66. What is the value of each animal ? 


6, A farmer bought 7 horses and 9 cows for $330. He could 
have bought 10 horses and 5 cows for the same money; find the 
price of each animal. : 


7, Twice A’s age exceeds three times B’s age by 2 years ; if 
the sum of their ages is 61 years, how old are they ? 


8, Half of A’s age exceeds a quarter of B’s age by 1 year, and 
three-quarters of B’s age exceeds A’s by 11 years; find the age of 
each ? . 


9, Ineight hours C walks 3 miles more than D does in six hours, 
and in seven hours D walks 9 miles more than C does in six hours ; 
how many miles dees each walk per hour ? 


10. In 9 hours a coach travels one mile more than a train does 
in 2 hours, but in 3 hours the train travels 2 miles more than 
the coach does in 13 hours ; find the rate of each per hour. 


11, A bill of $15 is paid with half-dollars and quarters, and 
three times the number of half-dollars exceeds twice the number of 
quarters by 6; how many of each are used ? 


12, A bill of $8.70 is paid with quarters and dimes, and five 
times the number of dimes exceeds seven times the number of quar- 
ters by 6; how many of each are used ? 


18, Forty-six tons of goods are to be carried in carts and wag- 
ons, and it is found that this will require 10 wagons and 14 carts, 
or else 18 wagons and 9 carts ; how many tons can each wagon and 
each cart carry ? 


14, A sum of $14.50 is given to 17 boys and 15 girls; the same 
amount could have been given to 13 boys and 20 girls; find how 
much each boy and each girl receives. 


15, A certain number of two digits is seven times the sum of 


the digits, and if 36 be taken from the number the digits will be 
reversed ; find the number, 


88 ALGEBRA. [ CHAP, XIV. 


16, A certain number of two digits is four times the sum of the ; 
digits, and if 27 be added to the number the digits will be reversed ; 
find the number. 


17, <A certain number between 10 and 100 is six times the sum 
of the digits, and the number exceeds the number formed by re- 
versing the digits by 9; find the number. 


18, The digits of a number between 10 and 100 are equal to 
each other, and the number exceeds 5 times the sum of the digits 
by 8; find the number. 


19, A man has $880 in silver dollars, half-dollars, and quarters ; 
the number of the coins is 852, and their weight is 235 ounces. If 
a dollar weighs 4 oz., a half-dollar + oz., and a quarter 4 oz., find 
how many of each kind of the coins he has. 


90, Aman has $22 worth of silver in half-dollars, quarters, and 
dimes. He has in all 70 coins. If he changed the half-dollars for 
dimes and the quarters for half-dollars, he would then have 180 
coins. How many of each had he at first ? 


91, Divide $100 between 3 men, 5 women, 4 boys, and 3 girls, 
so that each man shall have as much as a woman and a girl, each 
woman as much as a boy and a girl, and each boy half as much as 
aman and a girl. 


92, If 17 lbs. of sugar and 5 Ibs. of coffee cost $2.50, and 10 
lbs. of sugar and 10 lbs. of coffee cost $3.80, find the cost per lb. of 
sugar and coffee. 


93, The value of a number of coins consisting of dollars and 
half-dollars amounts to $22.50; the number of dollars exceeds five 
times the number of half-dollars by 6. Find the number of each. 


94, Asum of $23.80 is divided among 11 men and 16 women; 
the same sum could have been divided among 19 men and 6 women. 
Find how much each man and woman receives. 


95, Two articles A and B are sold for 20 cents and 30 cents 
per lb. respectively ; a person spends $6.50 in buying such articles. 
If he had bought half as much again of A and one-third as much 
again of B, he would have spent te 00. What weight of each did 
he buy ? 


CHAPTER XV, 


INVOLUTION. 


112, Deriyition. Involution is the general name for multi 
plying an expression by itself so as to find its second, third 
fourth, or any other power. 


Involution may always be effected by actual multiplication, 
Here, however, we shall give some rules for writing down at 
once 


(1) any power of a simple expression ; 
(2) the square and cube of any binomial ; 
(3) the square of any multinomial. 


113. It is evident from the Rule of Signs that 
(1) no even power of any quantity can be negative ; 


(2) any odd power of a quantity will have the same sign as 
the quantity itself. 


Note. It is especially worthy of remark that the square of every 
expression, whether positive or negative, is postive. 


114. From definition we have, by the rules of multiplication, 
Car a Oe he oe 
(— 28)? =(— a8 — 28) a8 =a, 
(—a°)? =(—a5)(—a®)( — a5) = — a5+545 = — Ql, 
(—307)?=(—3){a*)*=81a”, 
Hence we obtain a rule for raising a simple expression to any 
proposed power. 


Rule. (1) Ravse the coefficient to the required power by Arith- 
metic, and prefix the proper sign found by the Rule of Signs. 


) Multiply the index of every factor of the expression by the 
aes ve he power required. 


90 ALGEBRA. [cHap. 
Examples. (= 2277)8 S— Son" 
( —3ab?)§ = 729a5b}8, 


eer oa-b 


Ba°y/ = Sl ary 


It will be seen that in the last case the numerator and the denomi- 
nator are operated upon separately. 


EXAMPLES XV. a. 


Write down the square of each of the following expressions : 











eae, 9. Sac’. Soy" A.) +66%c% 
5, 4abc3. 6, —3a?y°. 1, =—2a7bd°c. 8, =—3da. 
azc Quz3 3a? 5 
a ith 12-75 
nO op ge mn® fe! _ 8pq?r?. 
iL. fates re. 9ay* 15, harry? 16. are 


Write down the cube of each of the following expressions : 


iW hk, | Pee 18, 3ab%. 19, 42°. 90, —3a’d. 

91, —427y7. 99 =bed?, 93, —-6y'*. 94, —4p%q° 
1 ee _ 82% iy 

25, pire 26. As a1. Ay 28. 3” : 


Write down the value of each of the following expressions : 


29, (ab?)4. 80, (-2?y)>. 31, (—2m2n’)6, 32, (—a3y2V’, 
mG) me CBR on CH we 


To Square a Binomial. 





115, By multiplication we have 
(a+6)?=(a+b)(a+b) 
07 200 Bs on. 5 ats ate eeoeee (1); 
(a—b)?=(a—b)(a—b) 
= P= 200 4-07. aa sche steals ae 


Xv. ] INVOLUTION. 91 


These formulz may be enunciated verbally as follows : 


Rule 1. The square of the sum of two quantities ts equal to 
the sum of their squares increased by twice their product. 


Rule 2. The square of the difference of two quantities is equal 
to the sum of their squares diminished by twice their product. 
Hxample 1. (2+ 2y)? = 274+2.%.2y+(2y)? 
= x7 + 4ay +4y*. 
Example 2. (2a3—8b?)?= (2a3)?—2 . 2a3 . 3b?+ (8b?)2 
= 4q°— 12a3b?+ 964. 


To Square a Multinomial. 
116, By the preceding article 
(a+b+c)?={(a+b)+cP 

=(a+b)?+2(a+b)e4+c? 

= a*+ b?+ 7+ 2ab+2ac +4 2be. 
In the same way we may prove 

(a—b+cP=a?+b?+ 0? —2ab+ 2ac— be. 
(a+b+c4+d)?=040+C4+d*+2ab4+2ac+ 2ad+ 2be+ 2bd + 2cd. 

In each of these instances we observe that the square con- 
sists of 

(1) the sum of the squares of the several terms of the given 
expression ; 

(2) twice the sum of the products two and two of the several 
terms, taken with their proper signs; that is, in each product 
the sign is + or — according as the quantities composing it 
have like or unlike signs. 

Note. The square terms are always positive. 

The same laws hold whatever be the number of terms in the 
expression to be squared. 


Rule. Zo find the square of any multinomial: to the sum of 
the squares of the several terms add twice the product (with the 
proper sign) of each term into each of the terms that follow tt. 


Ez. 1. («¢-—2y—382z)?=x? + 4y? + 92? --2.%.2y—-2.4.3824+2.2y.32 
= 207 + 4y? + 927 — day — 6az+ 12yz. 

Hix. 2. (1420 -—32?)?=1 4+ 40?+4 9at+2.1.27-2.1.382?-2.27.32? 
=1+427+ 924+ 4a - 6x? — 1223 
=]+4a — 2x? — 124° 4 9x4, 

by collecting like terms and rearranging. 


92 ALGEBRA. [CHAP. xv. 


EXAMPLES XV. b. 


Write down the square of each of the following expressions : 


apa. 27; Dae = 2y: 3, a+30. 4, 2a-3b. 
5, 38a+b. 6, x-—5dy. 7. 2n+7n. 8, 9-2. 
9, 2-ab. 10, abc+l. ll, ab-cd. 12, 2ab+ay. 
Tope = 22 14, 34+2pq. 15, x?-3a. 16, 2a-+ab. 
17, atb-e. 18, a-b-e. 19, 2a+b+¢. 
20, 2u-y-2 91, «x+3y—-2z. 90, we ea +1, 
93, 3x+2p-q. 94, 1-22-32. 95, 2-8u+22 


26, xty+a—). 27, m—n+p—4q. 28, 2a+8b+%—2y. 


To Cube a Binomial. 


117. By actual multiplication, we have 
(a+b) =(a+b)a+b)(a+6) 
= 03 +3a°b + 3ab? + b*. 
Also (a — 6b)? =a? —3a7b + 3ab? — Bb’, 
By observing the law of formation of the terms in these 
results we can write down the cube of any binomial. 
Example 1. (2a +y)3 = (2x)? + 3(2x)?y + 3(2ax)y? + y? 
= 8a? + 122°y + 6xy? + y?. 


Example 2. (8% - 2a?) = (38x) — 3(3x)?(2a*) + 3(8a)(2a?)? — (2a?)3 
= 272° — 542°a? + 36xa4 - 8a, 


EXAMPLES XV. c. 


Write down the cube of each of the following expressions : 

Le 2, m-n. 3, a- 2b. 4, 2c+d. 
5, x+3y. 6, «x+y. Ley: 8. 5442. 
Re rd be 10, 22?-+y7. ll, 2a%-30% 12, 4y?-3 


CHAPTER XVI. 


EVOLUTION. 


118, Derinirion. The root of any proposed expression is 
that quantity which being multiplied by itself the requisite 
number of times produces the given expression. 


The operation of finding the root is called Evolution: it is 
the reverse of Involution. 
119, By the Rule of Signs we see that 


(1) any even root of a positive quantity may be either positive 
or negative ; 

(2) no negative quantity can have an even root ; 

(3) every odd root of a quantity has the same sign as the 
quantity itself. 


Note. It is especially worthy of remark that every positive 
quantity has two square roots equal in magnitude, but opposite 
in sign. 


Hxample. J9a?x® = +3a2°, 
In the present chapter, however, we shall confine our attention to 
the positive root. 
Examples. ab! = 032, because (a%b2)? = a4. 
/ — x9 = — x, because (- 2°)® = - 2, 
Nc = ct, because (c4)> = c%, 
A/81a!2 = 3x3, because (323)! = S1x!2, 


120. From the foregoing examples we may deduce a general 
rule for extracting any proposed root of a simple expression : 

Rule. (1) Find the root of the coefficient by Arithmetic, and 
prefix the proper sign. 

(2) Divide the exponent of every factor of the expression by the 
index of the proposed root. 


Hxamples.  —642° = — 427, 


/ 16a8 = 2a2, 


/81a1° = 9x° 
25c8 Bc? 


94 ALGEBRA. [oHAP. 


EXAMPLES XVI, a, 


Write down the square root of each of the following expressions : 


{bap ese 2.6 20a Us 3, 49c7d%, A abe 
De Shaw ek 6, 162%. fis 0 avian Seopa 
4x6 ae isa. 144 
9, 16a4 10, 36° ag “O5 iby a2 
Write down the cube root of each of the following expressions : 
13, x*y®. 14, -a%®, 15, 82%, 16) 2/24 
pe bss 8a%h? 1250727) 64a77b? 
ive. 27° 18, “ys 19, ky [ae 20, ae nee 
Write down the value of each of the following expressions ; 
Dinan ey. DDN arta O38 N= ete 
24, N6ta®. 25, Na?tb™, 26, Npq”. 
97, N= x35y56, 298. N8laty®?. 29, §/320>bMe%, 


121. By the formule in Art. 115 we are able to write down 
the square of any binomial. 

Thus (20 + 3y)? =4u? + 12xy +97". 

Conversely, by observing the form of the terms of an expres- 
sion, it may sometimes be recognised as a complete square, and 
its square root written down at once. 


Example 1. Find the square root of 25a? — 40xzy + 16y’. 


The expression = (5a)?- 2. 20xy+ (4y)? 
= (5x)? — 2(5a)(4y) + (4y)? 
= (5x - 4y)?. 
Thus the required square root is 5x - 4y. 
64a? 32a 


Example 2, Find the square root of OOF UL Rae 


The expression = 3 i + (2)? + 2( 27 ) 


Thus the required square root is ee +2, 


XVI. ] EVOLUTION. 95 


122. When the square root cannot be easily determined by 
inspection we must have recourse to the rule explained in the 
next article, which is quite general, and applicable to all cases. 
But the student is advised, here and elsewhere, to employ methods 
of inspection in preference to rules. 


To Find the Square Root of a Compound Expression. 


123, Since the square of a+b is a?+2ab+b%, we have to dis- 
cover a process by which a and 0, the terms of the root, can be 
found when a?+2ab+0? is given. 

The first term, a, is the square root of a?. 

Arrange the terms according to powers of one letter a. 
The first term is a’, and its square root is a. Set this down as 
the first term of the required root. Subtract a? from the given 
expression and the remainder is 2ab +? or (2a+b) x 6. 


Now the first term 2ab of the remainder is the product of 
2a and b. Thus to obtain 6 we divide the first term of the 
remainder by the double of the term already found ; if we add 
this new term to 2a we obtain the complete divisor 2a+. 


The work may be arranged as follows : 


a’ +2ab+b? (a+b 
ae 


Qat+b 2ab +b? 
2ab+ 6? 


Example Find the square root of 9x? — 42xy + 49y. 





9x? — 42xy + 49y? ( 8a - Ty 
9x? 


6xu — Ty | —42xy + 49y? © 
| — 42xy +49y? 


Explanation. The square root of 92? is 3%, and this is the first 
term of the root. 


By doubling this we optain 6x, which is the first term of the 
divisor. Divide —42zy, the first term of the remainder, by 6x and 
we get —7y, the new term in the root, which has to be annexed both 
to the root and divisor. Next multiply the complete divisor by — 7y 
and subtract the result from the first remainder. There is now no 
remainder and the root has been found. 


96 ALGEBRA. [cHAP. 


124, The rule can be extended so as to find the square root of 
any multinomial. The first two terms of the root will be 
obtained as before. When we have brought down the second 
remainder, the first part of the new divisor is obtained by 
doubling the terms of the root already found. We then divide 
the first term of the remainder by the first term of the new 
divisor, and set down the result as the next term in the root 
and in the divisor. We next multiply the complete divisor by 
the last term of the root and subtract the product from the 
last remainder. If there is now no remainder the root has 
been found ; if there is a remainder we continue the process. 


Example. Find the square root of 
2527a? — 1220? + 1624 + 4a4 — 24270, 
Rearrange in descending powers of x. 
16a4 — 2443 + 25x70? — 12x03 + 4a4 ( 4a? - 8xa + 2a? 
16x4 


82? — 3x0 — 2420, + 25270? 
—24e2a+ 9x02 


82? — 6za + 2a? 1622a? — 12xa? + 4a4 
16z7a? — 1220? + 4a4 


Explanation. When we have obtained two terms in the root, 

4x? — 32a, we have a remainder 
1627a? - 12203 + 4a’. 

Double the terms of the root already found and place the result, 
8x2 —6xa, as the first part of the divisor. Divide 16xz°a?, the first 
term of the remainder, by 8x, the first term of the divisor; we get 
+ 2a? which we annex both to the root and divisor. Now multiply 
the complete divisor by 2a? and subtract. There is no remainder 
and the root is found. 





125, Sometimes the following method may be used. 


Example. Find by inspection the square root of 
4a? +b? +c¢?+4ab — 4ac — 2be. 
Arrange the terms in descending powers of a, and let the other 
letters be arranged alphabetically ; then 
the expression = 4a? + 4ab —4ac + b?- 2b¢4+ c? 
= 4a?+4a(b-c)+(b-c)? 
= (2a)7+2. Qa(b-c)+(b-c)?; 
whence tke square root is 2a+(b-c). [Art. 121.) 


XVI. ] EVOLUTION. 97 


EXAMPLES XVI. b. 


By inspection or otherwise, find the square root of each of the 
following expressions : 





1, @-8a+16. Q, «?+142+49. 

8, 6444824 922. 4. 25-30m+4 9m?. 

5, 36n4-84n?+49. 6, Sl+144y?+64y'%, 

T, 2° — 62% ytz4 + 9y8z8, 8, 4a2b4 — 12ab2c 4+ 919, 
ee ons 6 9a? , 24ac , 16c? 

9, ra 3xy? + 9y?. 10; yet bd +—a9- 
9a? 2562 16a? , 49y4 
Daa 0 all) a achat bd 2x4 

Nl. psp 2+ Gar 12. Foye" Tea oY 


18, 1624-3223 +242?- 82+]. 

14, 25 -30a+ 29a? - 12a°+ 4a4. 

15, 9a®-12a° - 2a4+4a?+1. 

16, 25p*- 30p? + 121 — 101p?+ 66p. 

17, 8v?+14+4a4- 42. 

18, 201a?- 108a? + 100 + 36a4 — 180a, 

19, a?+0?+c¢?+2ab —2ac — 2be. 

90, yet t+ 22x? + Py? — Qa?yz + Qary?z — Qarye. 


307 1 
4— q+ = 
Q1, at—2ab+ > - 5 +i5 
93, Imi+ 4 8 OO mtn 


94, 9at+ 144074 12ax?+ 4a? — 720° - 4802. 
95, x8 —4ax7 4+ dao + 62° — 1404+ 40° + 9a? -6r4+1. 
96, a?+9b?+c?- 6ab+6be —2ac. 


27, Mp crepe 47 +4, 


n* n? n? 


9a? b? 6a 2b. 
—-§4+—-— pe ze 
oF az 0b 


H.A. G 





98 ALGEBRA. [OnAr. 


{ If preferred, the remainder of this chapter may be postponed 
and taken at a later stage. ] 


To Find the Cube Root of a Compound Expression. 


126. Since the cube of a+b is a?+3a%b+3ab?+b3, we have 
to discover a process by which a and b, the terms of the root, 
can be found when a?+3a%)+3ab?+ 03 is given. 

The first term a is the cube root of a’. 

Arrange the terms according to powers of one letter a; then 
the first term is a, and its cube root a. Set this down as the 
first term of the required root. Subtract a’ from the given 
expression and the remainder is 

3a°b + 3ab? + D3 or (3a°?+3ab +b") x b. 

Now the first term of the remainder is the product of 3a? 
and 6. Thus to obtain b we divide the first term of the re- 
mainder by three times the square of the term already found. 

Having found b we can complete the divisor, which consists 
of the following three terms : 

1. Three times the square of a, the term of the root already 
found. 

2. Three times the product of this first term a, and the new 
term 0. 

3. The square of 6. 


The work may be arranged as follows : 





a +3ab+ 3ab?+b? (a+b 
he 
3(aPr = 3a? 3a°b + 3ab? +3 
3xaxb= +4+3ab 
Qe = +0? 





3a7+3ab+b? 3a°b + 8ab?+ 63 





Lxample 1. Find the cube root of 8x? — 36z°y + 54ary? — 27y?, 


8x3 — 36x?y + B4ay? — 27? ( Qa — 3y 
823 





B23)" = 109° 
3x 2x x (- 3y) = — 1ory 
(—3y)' = + 9y" 


1227 - 18xy + 9y? — 


— 86x7y + 54ay? — 27y? 






— 36x%y + 54ay? — 277 





va 


*PUNOF SI JOOL oY} PUL JopUIvUIeI OT 
sI rey} { ovagqns pue g— Aq JostAtp ozopdur0s oy Ajdynut Mon “g— jo orenbs oy4 ose puv “g— pue 
LF — LZ Jo qyoupoad oy} sow, g oye} 9M AOSTATP OT} aya{duroo OJ, *4OOL oy} JO W104 MoU B E— SOATS SIYy 
£ IOSLATP oY} Jo W199 4s1y oY) LTT Aq ‘AopuTVUTEL OY} Jo W194 4SIG OU} ‘,79E— SPLAT _“AOSTATP Mou Ot} jo 
qied si 0} Se “rept USF — VST “4[Nsert yy sovjd puv punos Apvarye joor oxy jo orenbs oy4 sourry g exe], 


*L]— XSOT — e206 — eX PFT 1 pXOE — 


JopuTeUTeI B OAT] OM “Zp — XZ “JOOI OY UT SUITE} OM} POUTeAGO OAvT OAL UOT] AA "wounUDnjidxy 

















Z 1G — LOT — -@O6 — eLFFI + X98 - 6 +298 + LOE + PSP — ASI 
eS 6+ =(¢—) 
RB LOE + UST - =(¢—) x (xp p#Z) x8 
iS LZ — X8OI — X06 — LEFT +4298 — LOb + COP — eTL = (ZH — eB) XE 
& = 

XF9 —pUOG + —USP—-| XII + ek FS— LVI 

erg + ‘ =(xp—) 
eXFG — = (xp—) x (2B) XE 
eX0S +5L09 + 6XSF - PXSL = o(zlG) XE 
9S 


2 -— XP— XLS) LZ — CSOT — X06 — eVOS8 + 5X09 t+ oT SP — 978 
‘1G — L8OL — e706 — eX08 + 4009 + oXSPF — 9B JO JOor oqno oy} PUNT °% apdumagy 


XVI. ] 


100 ALGEBRA. [CHAP. XVL 


EXAMPLES XVI. c. 
Find the cube root of each of the following expressions : 


], @+12a?+4+ 48a + 64. 9, 82? +120?+6xr+1. 
9 6423 — 1442+ 108a — 27. 4, 8p*-36p!+54p? - 27. 
5, m?—18m?+ 108m - 216. 6, x°+ 6xty? + 12a°y4 + 8y®. 
7, 1-3c+6c? — 7c? + 6c4 — 3c° + c%, 
8; 8+36m + 66m? + 63m? + 33m4 + 9m +m, 
Q, 216 -108k+ 342k? — 109K? + 171K4 — 27K + 27K8. 

10, 48y°+ 108y + 6044 — 90y? - 27 + 8y® — 80z°. 

11, 64+192k4 + 240k? + 160K? + 60K4 + 1245 + h°, 

12, x? —62°y — 3272+ 12xy? + lQxyz + 3x27 — 8y3 — 12yz - Byz? — 2. 


[For additional examples see Hlementary Algebra. | 


127, The ordinary rules for extracting square and cube roots 
in Arithmetic are based upon the algebraical methods explained 
in the present chapter. The following example is given to 
illustrate the arithmetical process. 


Hxample. Find the cube root of 614125. 


Since 614125 lies between 512000 and 729000, that is between 
(80)? and (90)°, its cube root lies between 80 and 90 and therefore 
consists of two figures. 

at+b 
614125 ( 80+5 = 85 
512000 


3a7 =.3 K (80) = 19200) 1102125 
3xaxb=3x80x5= 1200 
A= Bx Os 25 


20425 





In Arithmetic the ciphers are usually omitted, and there are 
other modifications of the algebraical rules. 


CHAPTER XVII. 


RESOLUTION INTO FACTORS. 


128. Derritiox. When an algebraical expression is the 
product of two or more expressions each of these latter quanti- 
ties is called a factor of it, and the determination of these 
quantities is called the resolution of the expression into its 
factors. 

In this chapter we shall explain the principal rules by which 
the resolution of expressions into their component factors may 


be effected. 


Expressions in which Each Term is divisible by a Common 
Factor. 


129. Such expressions may be simplified by dividing each 
term separately by this factor, and enclosing the quotient within 
brackets ; the common factor being placed outside as a coefficient. 


Example 1. The terms of the expression 3a? — 6ab have a common 
factor 3a; 
3a? -- 6ab = 8a(a — 2b). 


Example 2. 5a*ba? — 15aba? — 20032? = 5bx?(a*a — 3a — 407). 


EXAMPLES XVII. a. 


Resolve into factors : 


], 2+ax. Oy 262 =. a0. Ce Os. 4, ai--a?b. 
5, 3m2-6mn. 6, p?+2p?q. T, 2-52, 8, yxy. 
Q, 5a? -—25a7b. 10, 12%+4827y. ll, 10c? — 25c4d. 
12, 27-162z. 13, x°y?+82y. 14, 17x2-5la. 
15, 2a?-a?+a. 16, 32°+6a?x? - 3a°x. 
17, 7p?-7Tp? + 14p%. 18, 40°+6a7b? — 262. 


19, xy? — 2?y?+ 2Qary. 920, 26a°b°+39a%b2, 


102 ALGEBRA. [onar. 


Expressions in which the Terms can be so grouped as to 
contain a Compound Factor that is Common. 
180, The method is shown in the following examples. 


Example 1. Resolve into factors «2—ax+bxe—ab. 


Since the first two terms contain a common factor x, and the 
last two terms a common factor b, we have 


x? ~— ax +ba—ab = (x?—ax)-+(ba-ab) 

a(x ~a)+b(~-a) 

(a —a) taken w times plus (w-—a) taken 0 times 
= (x-—a) taken (x+b) times 

= (~-—a)(~+b). 


II 


I 


| EHxample 2. Resolve into factors 6a? — 9ax + 4ba - Gab. 
6x? — 9ax+4bx - bab = (6x? — 9ax) + (4ba — 6ab) 
= 3a(2a —- 3a) + 2b(2x - 3a) 
= (24 — 3a)(3a” + 2b). 
Example 3. Resolve into factors 12a? + ba? — 4ab — 3ax. 
12a? + ba? — 4ab — 38ax? = (12a? — 4ab) — (Bax? — bx?) 
¢ = 4a(3a —b)- x°(8a —- b) 
= (3a —b)(4a - x). 


EXAMPLES XVII. b. 

Resolve into factors : 
1, x?+ayt+ue+y% 2, w—xz+auy-y2. 
8, a?+2a+ab+2b. 4, a®?+ac+4a+4e. 
By Qa+2xe+an+x2, 6, 3q-3p+pq-p. 
7, am—bm-an+bn. 8, ab—by-ay+y’. 
9, pataqr-pr-7. 10, 2ma+na+2Qmy + ny. 
ll, awv-2ay - ba +2by. 12, 2a?+3ab —2ac — 3le. 
18, ac?+b+bc?+a. 14, ac*—2a -bc?4+ 2b. 
15, a@-a?+a-l1. 16, 22°+3+4+2a+4 32. 
17. ax—aby+2ax—2by. 18. axy+bexy—az—bez. 


Trinomial Expressions. 


181, In Chap. v. Art. 48 attention has been drawn to the 
way in which, in forming the product of two binomials, the 
coefficients of the different terms combine so as to give a trino- 
mial result. 


XVII. ] RESOLUTION INTO FACTORS. 103 


Thus (BED GAS) Sa FBG ALD, ets scse veces (1), 
(a —5)\ae— et HO Arty Le cain cea ina ies cath pae (2), 
CoD BN OD cciaces voden se Catees eee 
CoD Reto a 2b LO ans cein ew saddens (4). 


We now propose to consider the converse problem: namely, 
the resolution of a trinomial expression, similar to those which 
occur on the right-hand side of the above identities, into its 
component binomial factors. 


By examining the above results, we notice that : 
1. The first term of both the factors is 2. 


2. The product of the second terms of the two factors is 
equal to the third term of the trinomial; e.g. in (2) above we 
see that 15 is the product of —5 and —3; ‘and in (3) we see 
that —15 is the product of +5 and —-3. - 


3. The algebraic sum of the second terms of the two factors 
is equal to the coefficient of # in the trinomial ; e.g. in (4) the 
sum of —5 and +3 gives —2, the coefficient of w in the tri- 
nomial. 


The application of these laws will be easily understood from 
the following examples. 


Example 1. Resolve into factors v2+ lla +24. 


The second terms of the factors must be such that their product 
is +24, and their sum +11. It is clear that they must be +8 
and +3. 

x? + 1la+24 =(x+8)(~+4+3). 

Example 2. Resolve into factors x? - 10x” +24, 


The second terms of the factors must be such that their product 
is +24, and theirsum —10. Hence they must both be negative, and 
it is easy to see that they must be -6 and —4. 


x? —10x”+24 = (a -—6)(a - 4). 


EHxample 3. x?-18x%+81 = (x-9)(x-9) 
= (x - 9). 

Example 4. + + 10a? + 25 = (2? + 5)(y?+5) 
ea (ro Ve 


Example 5. Resolve into factors x? - llaa+10a?. 
The second terms of the factors must be such that their product 
is +10a7, and their sum —lla. Hence they must be —-10a and —a. 
—llax+10a? = (x - 10a)(x — a). 
Note. In eed: of this kind the student should always verify 


his results, by forming the product (mentally, as explained in 
Chap. v.) of the factors he has chosen. 


104 ALGEBRA. [CHAP. 


EXAMPLES XVIL ec. 
Resolve into factors: 
ik +3e+ 


2. 2. y+5y+6. 3. x tiy+12. 
4. a?—3a+2. 5. a?—6a+8. 6, B-55+6. 
7, +135+42. 8. 6-135+40. 9, =-132+36. 
10, 2°-15r+56. ll, 2?-157454. 12, 24+152+44. 
13, 126436. 14, a@84+15a+56. 15, a?—120+27. 
16, 2°+92+20. 17, z2-10r+9. 18, 2°-162+64. 
19, »°-23y+102. 90, y®-24y+95. 21. 2+ 54y +729. 
99. a-+10ab+21F%. 23, a? +12ab+11F. 24, a? - 23064 132%. 
95, m*+8m?+7. 26. 429m*n?+14n*. 27, 6-S5rix. 


98, 54-l5a+a?. 29. isp 30, 216-35a+a*. 


1382. Next consider a case where the third term of the tri- 
nomial is negative. 


Example 1. Resolve into factors z*+2z—- 35. 


The second terms of the factors must be such that their product 
is —35, and their algebraical sum +2. Hence they must have 
opposite signs, and the greater of them must be positive in order 
to give its sign to their sum. 

The required terms are therefore +7 and —5. > 

x? + 2a —35 = (x+7)(x-5). 


Example 2. Resolve into factors z* — 3z - 54. 


The second terms of the factors must be such that their product 
is —54, and their algebraical sum —3. Hence they must have 


opposite signs, and the greater of them must be negative in order 
to give its sign to their sum. 


The required terms are therefore —9 and +6. 
x? -3z2-54 =(x-9)x+6). 

Remembering that in these cases the numerical quantities 
must have opposite signs, if preferred, the following method may 
be adopted. 

Example 3. Resolve into factors z*y* + 23zy — 420. 


Find two numbers whose product is 420, and whose difference is 
23. These are 35 and 12; hence inserting the signs so that the 
positive may predominate, we have 


ry"? + 23zy — 420 = (zy + 35)(zy — 12). 


xviIL] RESOLUTION INTO FACTORS. 105 
EXAMPLES XVIL @ 
Resolve into factors: 
1, 242-2 2. -z-6. 3, 2-2-2. 
4. y+4y-12 5. y+4y-21. 6. y°-5y-36. 
7, @+Sa-3. 8. 2-14-30. 9, wia- 132. 
10. --—12%-4. lL &+145-51. 12. F+10-39. 
13, m*-—m—56. 14, m-im-S. 15, minm- 5. 
16. ~-—Sp-6&. 17. #+3p-108. 18. #+p-110. 
19, 2+27-4. 90, = -—7z-120. 91. =-2z-132 
92. xt+139¢-48. 23. t4dzry-S9. 24. ¥+izy-Ke. 
% aia FT 96. a+ab-—240F. o7, 14-i-&. 
28. 3-%-¥. 29. 9645-1. 90, 72+5-F 


133. We proceed now to the resolution into factors of tri- 
nomial expressions when the coefiident of the highest power is 
not unity. 

Again, referring to Chap. v. Art. 48, we may write down the 
following results - 


se et Oe A, (1), 
— 2) x—4}=32°— 1448.0... (2), 
ons (x—4)=322— eoige hee eee, (3), 
—9\(x+-4)=32"+410e—8........---------- (4) 


The converse bea fe presents more ORS than the cases 
we have yet considered. 

Before endeavouring to give a general method of procedure 
it will be worth while to examime im detail two of the identities 
given above. 


Consider the result 32° — 14r+8=(3r—2yYxr—4)} 
The first term 32° is the product of 3r and x. 
The third term +8....................- —land -4+4 


The middle term —14r is the result of adding together the 
two products 3r x —4 and xx —2 

Again, consider the result 3x.°*—10r—8=(3r+2\2-—4) 

The first term 3 is the product of 3r and x. 

The third term —8..................... +2and -+ 

The middle term — 1 ts the result of adding together the 
two products 3rx —4and 2x2; and its sign is negative because 
the greater of these two products is negative. 


106 ALGEBRA. (CHAP. 


134, The beginner will frequently find that it is not easy 
to select the proper factors at the first trial. Practice alone will 
enable him to detect at a glance whether any pair he has chosen 
will combine so as to give the correct coefficients of the expres- 
sion to be resolved. 


Example. Resolve into factors 7x? - 19x - 6. 


Write down (7z 38)(@ 2) for a first trial, noticing that 3 and 2 
must have opposite signs. These factors give 7x? and —6 for the 
first and third terms. But since 7 x2—3x1=11, the combination 
fails to give the correct coefficient of the middle term. 


Next try (7x 2)(a 3). 


Since 7x38-2x1=19, these factors will be correct if we insert 
the signs so that the negative shall predominate. 


Thus 7a? — 192-6 = (7x +2)(x — 8). 
[Verify by mental multiplication. ] 


1385, In actual work it will not be necessary to put down 
all these steps at length. The student will soon find that the 
different cases may be rapidly reviewed, and the unsuitable 
combinations rejected at once. 

It is especially important to pay attention to the two follow- 
ing hints : 


1. If the third term of the trinomial is positive, then the 
second terms of its factors have both the same sign, and this 
sign is the same as that of the middle term of the trinomial. 


2. If the third term of the trinomial is negative, then the 
second terms of its factors have opposite signs. 


Example 1. Resolve into factors 14x%?+ 29% -15 ....... ee, (1) 
]474— 203-15 se eee (2). 


In each case we may write down (72 ~3)(2x% 45) as a first trial, 
noticing that 3 and 5 must have opposite signs. 


And since 7 x 5-3 x 2=29, we have only now to insert the proper 
signs in each factor. 


In (1) the positive sign must predominate, 

Ati A ZY UNS NEGA UY Cr...:-cas+s ccs aecrest aeee eye 

Therefore 1427+ 29a —15 = (7a — 3)(2a +5). 
7 


XVII. J RESOLUTION INTO FACTORS. 107 


Example 2. Resolve into factors 547+ 172 +6 .....cccccceccseeee eee (1); 

arid PEA Oe tan tis eae coo (2). 
In (1) we notice that the factors which give 6 are both positive. 
EDC ys Sis oan sen ie a aC Oe a asc negative. 
And therefore for (1) we may write (5v+ )(x+ ). 


And, since 5x 3+1x2=17, we see that 
5a? +172+6 = (5a+2)(x+3). 
52? — 17246 = (5a —2)(x- 3). 


Note. In each expression the third term 6 also admits of factors 
6 and 1; but this is one of the cases referred to above which the 
student would reject at once as unsuitable. 


EXAMPLES XVII. e. 


Resolve into factors : 


1, 2a7+3a+1. 9, 3a7+4a+4+1. 3, 407+5a4+1. 
4, 2a?+5a4+2. 5, 3a7+10a+3. 6, 2a?+7a+3. 
7, 5a7+7a+4+2. 8, 2a7+9a+10. Q, 2a?+7a+6. 
10, 2%7+92%+4+4. 11, 2%?+52 - 3. 12, 327+52-2. 
13, 39?+y-2. 14, 3y?-7y-6. 15, 2y?+9y-5. 
16, 207-5b-3. 17, 60?+7b-3. 18, 26°+b-15. 


19, 4m?+5m-6. 90, 4m?-4m-38. Q1, 6m?-7m-3. 
99, 427-8xry—-—5y*. 98, 6x?-Tayt+2y*, 94, 6a?-1382y+2y?. 
95, 12a?-17ab+6b?. 26, Ga?-5ab-6b7. 97, Ga?+35ab— 607. 
28, 2-3y- 2y*. 29, 3+23y -8y?. 80, 8+18y—5y2. 
9], 44172-1522. 39. 6-13a+6a%. 33, 28-31b—502 


When an Expression is the Difference of Two Squares. 
186. By multiplying a+b by a—b we obtain the identity 
(a+b)(a—b) =a?—}?, 
a result which may be verbally expressed as follows: 


The product of the sum and the difference of any two quantities 
is equal to the difference of their squares. 

Conversely, the difference of the squares of any two quantities 
is equal to the product of the sum and the difference of the two 
quantities. 

Thus any expression which is the difference of two squares 
may at once be resolved into factors. 


108 ALGEBRA. [CHAP, 


Example. Resolve into factors 25x? - 16y?. 
25a? — 16y? = (5x)? — (4y)?. 
Therefore the first factor is the sum of 5% and 4y, 
and the second factor is the difference of 52 and 4y. 
2527 — 16y? = (5a + 4y)(5ax — 4y). 
The intermediate steps may usually be omitted. 
Example. 1 — 49c® = (1+ 7c3)(1 — 7°). 


The difference of the squares of two numerical quantities is 
sometimes conveniently found by the aid of the formula 


a? —b?=(a+b)(a—b). 


Example. (329)? — (171)? = (329 + 171)(329 - 171) 
= 500 x 158 
= 79000. 


EXAMPLES XVII. f. 


Resolve into factors : 


1, a’-9. 9, a? -49. 3, a’-8l. 4, a’?-100. 
Boa = 25. 6, «144. 7. 64=:2. 8, 81-42". 
Q, 4y?-1. 10, y?-9a% Il, 4y°-25. 12, 9y?-49z%. 
18, 4m?-81. 14, 36a?-1. 15, k?-640?. 
16, 9a?- 250". L]7, (121 1697, 18, 121-3627. 
LON, 25 =e 90, a?b? — a7y". 91, 49a*— 1006. 
992, 64a? - 492. 93, 4p?q?-81. OA rat hea a D 
95, 2x®- 4a, 06, x7 = 2027. 1 0 = po. 
98, 16a!6- 90%. 99, 2xa’-4, 380, a%b8c4 — 92. 
Find by factors the value of 
Ql, (89)? — (31). 82, (51)? - (49). 33, (1001)?-1. 
34, (82)-(18)?. 35, (275)?- (225), 36, (936)? — (64). 


When an Expression is the Sum or Difference of Two Cubes. 

187. If we divide a?+b3 by a+b the quotient is a?—ab+6?; 

and if we divide a?—b3 by a—b the quotient is a?+ab+b”. 

We have therefore the following identities : 
a®+b3=(a+b)(a?—ab+b?) ; 
a®—b3=(a—b)(a?+ab+b?). 

These results enable us to resolve into factors any expression 

which can be written as the sum or the difference of two cubes. 


XVII. | RESOLUTION INTO FACTORS. 109 
Hxample 1. 8a? — 27y? = (2x)? - (By)3 
= (2a — 3y)(4u? + Gay + 9y?). 
Note. The middle term 6zy is the product of 2a and 3y. 
Example 2. 64a? +1 = (4a)? +(1)3 
= (4a + 1)(16a2 — 4a +1). 


We may usually omit the intermediate step and write down 
the factors at once. 


Examples. 343a® - 272° = (7a? — 3x)(49a4 + 21a? + 92"). 
8x9 +729 = (223+ 9)(4a% — 1823 +81). 


EXAMPLES XVII. g. 


Resolve into factors : 


ec = U2, 9. a +b°. Bala: 4, 1-7. 
5, S8a?+1. 6, x? — 823. Le 0 270°. 8, 2-1. 
9, 1-8a?. 10, 08-8. A ant bee 12, 64-p*. 

13, 12503+1. 14, 216-23 15, xy? +343. 

16, 1000a°+1. ie Silas 7. 18, a*b°®c? — 27. 

19, Sx? - 343. 90, x? +2167". 91, x8 -— 272. 

92, m—1000n°. 93, a? —729b%, 94, 125a%+512b%. 


138, We shall now give some harder applications of the 
foregoing rules, followed by a miscellaneous exercise in which 
all the processes of this chapter will be illustrated. 


Example 1. Resolve into factors (a + 2b)? — 162°. 
The sum of a+ 20) and 4x is a+2b+ 4a, 
and their difference is a+2b-—- 4a. 
(a +2b)? — 16x? = (a+2b +4x)(a+2b—4zx). 


If the factors contain like terms they should be collected so 
as to give the result in its simplest form. 


Example 2. (8”+7y)? — (2x -3y)? 
= {(32+ Ty) + (2a - 8y)} {(3a-+Ty) - (2x — 3y)} 
= (8a¢+7Ty + 2x - 3y)(8x+ Ty — 2a + 3y) 
= (5¢+4y)(~+10y), 


110 ALGEBRA. [ CHAP. 


139. By suitably grouping together the terms, compound 
expressions can often be expressed as the difference of two 
squares, and so be resolved into factors. 


Hxample 1. Resolve into factors 9a? — c?+4cu - 4a”, 
9a? — c? + 4ca — 4a? = 9a? — (c? — 4ca + 477) 
= (3a)? — (c — 22)? 
= (8a+c¢ — 2x)(3a —-c+2z2). 


Example 2. Resolve into factors 2bd — a? — c? + b?+ d?+2ac. 


Here the terms 2d and 2ac suggest. the proper preliminary 
arrangement of the expression. Thus 


2bd — a? —c? + b? + d*+2ac = b?+ 2bd + d? — a? + 2ac — c? 
= b?+ 2bd + d? — (a? — Qac +c?) 
= (b+d)?-(a-c)? 
=(b+d+a-c)(b+d-atec). 


140. The following case is important. 


Example. Resolve into factors 24+ 22y? + y4. 
x4 ee al ab y4 = (a*+ Dy224 2 +y') ms raf 
= (22+ 92)? = (ay)? 
= (2? -+y? + ay)(a?+y? — xy) 
= (x? +ay+y?)(a?-xy+y?). 


141, Sometimes an expression may be resolved into more 
than two factors. 


gf 


£1 1 


Example 1. Resolve into factors 16a‘ -— 814. 
16a4 — 81b4 = (402+ 9b2)(4a2 — 9b?) 
= (4a? + 9b*)(2a + 3b)(2a — 3b). 


Example 2. Resolve into factors 2° — y°, 
af — yf = (a? + y°)(28 — y?) 
= (e+ y)(a?— ay + yu —y)(a?+ay ty), 
Note. When an expression can be arranged either as the dif- 
ference of two squares, or as the difference of two cubes, each of the 
methods explained in Arts. 136, 137 will be applicable. It will, 


however, be found simplest to first use the rule for resolving into 
factors the difference of two squares. 


XVII. | 


RESOLUTION INTO FACTORS. 


111 


142, In all cases where an expression to be resolved contains 
a simple factor common to each of its terms, this should be first 
taken outside a bracket as explained in Art. 129. 


Huample. 


Resolve into factors 28x4y + 64a°y — 60x7y. 


28u4y + 642%y — 60x?y = 4x?y(7x? + 16x — 15) 
= 4a°y(7x — 5)(@ +3). 


EXAMPLES XVII. h. 


Resolve into two or more factors : 


1, (w+y)-2. 2, (w-y)-2. 

4, (a+3c)?-1. 5, (2¢-1)?-a?. 

7, 4a?-(b-1)?. 8, 9-(a4+2). 
10, (18%+y)?-(17"-y). its 
12, 4a? - (2a - 3b). 13. 
14, (x+y)? (m-n)?. 15. 
16, a@-2ax%+a?- 4b. live 
18, 1-a?—2ab-—b?. 19. 
90, c?-a?-b?+2ab. O11. 
99, x*+y*—2t-at4 Qary? —2Qa72?, 93, 
94, at+a?+l. 95, a‘b*—16. 
27, 16a4b?- 6%. 98, 64m7 — mn, 
80, «@7b>-81a?d. 81, 400a?x — x, 
83, 2160° + ad, 34, 25027 +2. 
386, aa? — ax? -240ax. 37, 
88, mt+4m?n2p? + 4ntp+. 39, 
40. Gay? + 1l5a°y? - 36ay". 41, 
42, 9824 — Tay? — yy? 48. 
44, 2? -2u7-4%+4+2. 45, 
46, a’2x? — 8a2y? — 4b7x3 + 32077". 
47, 2p-3q+4p? -9q°. 48, 
49, 24a7b? — 30ab? — 3604. 50, 
HI, 2*+427+4+ 16. bo: 
58, a*t—18a7b? + b+. 54, 


8, (a+2b)? -—c?. 

6, @-(b+c). 

Q, (2a-—3b)?-c3. 
(6a + 3)? — (5a — 4)”. 


x — (2b -- 8c)". 

(3x + 2y)? — (2a - 3y)?. 

x? + a? + Qaa — 22, 

12ay +25 — 4x7 - 9y?. 

x? — Qa +1 —m?—-4mn —4n2. 
(m+n+p)?—-(m-n+p). 

96, 25624 -81y%. 

99, at—aty?. 

ou. L—=129y*. 

95, 1029—3x% 
aca? + bex — adx — bd. 
82243 — 20°. . 
2m8n*t — Tmin’ — 4n8, 
a°b? — a? — 6? +1. 
(a@+bpP+1. 


119+ 10m — m?. 
240? + aSy/4 — a7l0y8, 
at + y+ — Ta2y?, 

8 +o4+1. 


[For additional examples see Elementary Algebra. ] 


112 ALGEBRA. [crraP. 
Converse Use of Factors. 


148, The actual processes of multiplication and division can 
often be partially or wholly avoided by a skilful use of factors. 


It should be observed that the formulze which the student 
has seen exemplified in this chapter are just as useful in their 
converse as in their direct application. Thus the formula for 
resolving into factors the difference of two squares is equally 
useful as enabling us to write down at once the product of the 
sum and the difference of two quantities. 


Hxample 1. Multiply 2a+3b-c¢ by 2a-3d+¢. 


These expressions may be arranged thus: 
2a+(8b-—c) and 2a-—(3b-c), 
Hence the product = {2a + (3b —c)} {2a — (3b —c)} 
= (2a)? - (8b -c)? 
= 4a? — (9b? — 6bc +c?) 
= 4a? — 9b? + 6be - c?. 


EHxample 2. Find the product of 
x+2, 2-2, w?-Qa+4, u?+2r+4. 
Taking the first factor with the third, and the second with the 
fourth, 
the product = {(a + 2)(x? — 2a +4)} {(a - 2)(u?+2x%+4)} 
= (a + 8)(a — 8) 
= 26 — 64. 


Example 3. Divide the product of 2u?+a”-6 and 62?-5x+1 
by 3x7+5x-2. 

Denoting the division by means of a fraction, 
(2a?+ x2 — 6)(6x? —5x”+1) 

327 +52 —2 
(2a — 8)(a +2)(3x —1)(2a—1) 
(8% —1)(%+2) 
= (2% —3)(2x- 1), 

by cancelling factors which are common to numerator and denomin- 
ator. 


the required quotient = 





XVII. ] CONVERSE USE OF FACTORS. TS 


Example 4. Prove the identity 
17(5x + 3a)? — 2(40a + 27a)(5x + 3a) = 25x? - 9a. 
Since each term of the first expression contains the factor 5x + 3a, 
the first side = (5% + 3a){17(5x + 8a) — 2(40x + 27a)} 
= (5x + 8a)(85x + 5la — 80x — 54a) 
= (5% + 3a)(5x — 3a) 
= 252? — 9a?. 


EXAMPLES XVII. k., 


Employ factors to obtain the product of 


1, a-b+c, a-b-e. 2, 2w—yt2, W+y+z 

8, 14+2x-x7, 1-2x-2?, 4, ¢c?+3c4+2, c?-3c-2. 

5, a+b-—c+d, at+b+c-d. Go Po rey, p-g— ary. 
7, a’®—4a2d + 8ab? —-8b°, a*+4a°b + 8ab? + 8b? 


Find the continued product of 


8, (a—b), (a+b), (a? +b?) 
9, (icy, (L+2)°, (ia? )". 
10, a-4a+3, a?-a-2, a?+5a+6. 
ll, 3-y, 3+y, 9-3y+y’, 94+3y+y’. 
12, l+ct+e?, l-c+e, 1-c?+c4 
18, Divide a3(a+2)(a*-a-56) by a?+7a. 
14, Divide the product of z7+a”-2 and 27+4x%+3 by 2?+5x+6. 
15, Divide 32°(2+4)(z?-9) by 2?+a”-12. 
16, Divide the product of 2x%7+lla-21 and 38a?-20a-7 by 
a” — 49. 
17, Divide (2a?-a-3)(8a?-a-2) by 6a7-5a—-6. 
18, Divide x6-7x?-8 by (~#+1)(u?+2u +4). 


Prove the following identities : 


19, (a+b)?-(a-b)?(a+b) = 4ab(a+b). 

90, ct-—dt—(c—d)(c+d) = 2cd(c? - d?). 

91, (m-n)(m+n) — m+ nt = 2mn(m? — v7). 
22, (a+y)t-8ay(a+y)? = (a+y)(a+y"). 
93, 3ab(a—b)?+(a—b)* = (a —- b)(a? — b*). 


H.A. vee 


114 ALGEBRA. (CHAP. 


MISCELLANEOUS EXAMPLES III. 


], Find the product of 10z?-12-3% and 2%-4+432%. 

Qo Leal. b= 1 ea 2, 0.0, tindsthe walle of 
a?—b?  b*-cd , cb - 0? 
a2+b? 2b?+cd Babe° 

3, Simplify 2[4a - {2y+(2e-y)-(«+y)}]. 

4, Solve the equations : 

(1) z—-3 3-9 12x (2) 3u — 4y = 25, 
5 3 15 ie bt 2y =i. 
5, Write down the square of 223-—a+5. 
6, Find the H.C.F. and L.C.M. of 3a2b°c, 12a4b?c?, 15a%b°c. - 








7, Divide a*+4b* by a?—2ab+26?. 
8, ind in dollars the price of 5x articles at 8a cents each. 
9, Find the square root of 24-82? + 24a? — 32x+16. 
QO, lh.a=5, b=3.¢ =]; ind the value of 
(a—b) , (b-c) , (ac) 
a+b b+e a+c¢ 
Ti eeccire (78 +5) -~73=13 -f(a i AI. 


12, A is twice as old as B; twenty years ago he was three 
times as old. Find their ages. 





18, Simplify (1 - 2a) -{3-(4-5x)}+{6- @7 -8z)}. 
14, The product of two expressions is 
6x4 + Say + 6x?y? + Bay? + 6y4, 
and one of them is 247+ 3ay+2y?; find the other. 


15, How old is a boy who 2x years ago was half as old as his 
father now aged 40? 


16, Find the lowest common multiple of 2a?, 3ab, 5a%bc, 6ab*c, 7a”. 
17, Find the factors of 
(1) 2?-axy-72y*. (2) 6x?-18x%+4+6. 
18, Find two numbers which differ by 11, and such that one- 
third of the greater exceeds one-fourth of the less by 7. 


XVII. ] MISCELLANEOUS EXAMPLES III. 115 


ifo@al, b=—)) c= 2, d.=0, find.the value of 


at+b c+d, ad— be een 
a-b ce-d bd+ac a?+b™ 


- : 3 ae | ad be 
90, Simplify 8x —y—{2v-ly-7-(%-4)+(2-}e)l 


91, Solve the equations : 
(1) (3a -8)(8x% +2) -— (4% -—11)(2%+1) =(%-3)(x%4+7) 5 
x — ; 25 ene 
(2) SU Siva 2, aty-5=ay-2). 
99, <Atrain which travels at the rate of p miles an hour takes 
q hours between two stations; what will be the rate of a train 
which takes 7 hours? 
93, Find the sum of 
eel a 2 ie Fete | 
re es, ]1- 5 ac (20-5), Pele vite 


19, 











24, Resolve into factors 
(1) 12x?+ax—20a? ; (2) a?-16-6axr+9x2, 
25, Solve 


(1) x+1+2(%+38) = 4(~+5); 
(2) 4%+9y=12, 6x-3y =7. 
: —x+ V3 = 222 aul 
96, Find the value of AMEE when 2 = 3" 
97, Find the quotient wanes the product of b?+c? and b?-¢% 


is divided by b? — 2b*c + 2bc? - c?. 
98, A, B, and C have $168. between them; A’s share is greater 
than B’s by $8, and C’s share is three- fourths of A’s. Find the 


share of each, 
99, Find the square root of 9x° - 
30, Simplify by removing brackets a? —[(b -c)* - {c? - (a —b)?}}. 





122° + Q2at + a? + 1994 4, 


CHAPTER XVIII 


HIGHEST COMMON FACTOR. 


144, Derrimirion. The highest common factor of two or 
more algebraical expressions is the expression of highest dimen- 
sions which divides each of them without remainder. 


Note. The term greatest common measure is sometimes used 
instead of highest common factor ; but this usage is incorrect, for in 
Algebra our object is to find the factor of highest dimensions which 
is common to two or more expressions, and we are not concerned 
with the nwmerical values of the expressions or their divisors. The 
term greatest common measure ought to be confined solely to arith- 
metical quantities, for it can easily be shown by trial that the 
algebraical highest common factor is not always the greatest 
common measure. 


145, We have already explained how to write down by 
inspection the highest common factor of two or more simple 
expressions. {See Chap. x11.] An analogous method will enable 
us readily to find the highest common factor of compound ex- 
pressions which are given as the product of factors, or which 
can be easily resolved into factors. 


Example 1. Find the highest common factor of 
4cx3 and 2cx?+4c?x*. 
It will be easy to pick out the common factors if the expressions 
are arranged as follows : 
Aen = sen, 
2cx8 + 42a? = Wcx?(a+2c) ; 
therefore the H.C.F. is 2ca?. 


Hxample 2. Find the highest common factor of 
3a7+9ab, a*®—9ab?, a? +6a2b + 9ab. 
Resolving each expression into its factors, we have 
3a7+9ab = 8a(a+ 3b), 
a® — 9ab? = a(a+3b)(a— 3b), 
a? + 6a*b + 9ab? = a(a+3b)(a+3b) ; 
therefore the H.C.F. is a(a+3b). 


CHAP. XVIII. ] 


HIGHEST COMMON FACTOR. 


fees 


146, When there are two or more expressions containing 
different powers of the same compound factor, the student should 
be careful to notice that the highest common factor must contain 
the highest power of the compound factor which is common to 
all the given expressions. 


Example 1. 


The highest common factor of 


x(a-<x), a(a—x)*, and 2ax(a—x)* is (a-x). 


Example 2. Find the highest common factor of 
ax? +2a?x+a3, 2ax*- 4a7?x - 6a, 3(axe+a?)’. 


Resolving the expressions into factors, we have 


ax? + 2a7% +a? = a(x? +2aa +a?) 


=a(x+a)? 


Seer ne eres soe eee eeeeeress sed 


Qan? — 4a7x — 6a? = 2a(a? — 2a - 3a?) 


= 2a(x~+a)(x - 3a) 
3(ax +a”)? = 3f{a(x+a)}? = 8a%(x+a)? 


(3). 


Ce ey 


Therefore from (1), (2), (3), by inspection, the highest common 
factor is a(x+a). 


Find the highest common factor of 


on oo 


11. 
13, 
15. 
ING 
19. 
21, 
22, 


EXAMPLES XVIII. a. 


x7 —y", x — ay. 2. 
34° —2a7b, 3a? - 2Qab. 4, 
ct — cd, c4— cd?. 6. 
a’a7(a — 2), Qa2x?(a — x). 8. 
ax+2, atv+an. 10. 
xr+ay", x?+ay, x°y+ay, 19, 
(@?-ax)’, (ax—2°*)?, 14, 
x? —42— 42a, 24-4922, 16. 
a® — 36a, a?+2a?— 48a, 18. 


Qu? —9x+4, 3u2-7xr-20. 20, 


3(a —b)3, a®-2ab4+ 0%. 

9a? —4b7, 6a*+4ab. 

an — atty?, ay? + 2273, 

Qa? — 8248, (x-2)% 
ayrt— yt, xy ty’, zy-y’. 
vy—y’, y(ey-y’). 
(abe — bc*)?, (ac — ac?) 
(a? — 5a?)?, a° — Sat+ 1523. 
3a7+7a-—6, 2a?+7a+3. 
3c4+5c3 —12c?, 6c> + 7c4 — 20c?. 


9 
a 
° 


4m —9m?, 6m? —5m?-6m, 6m4+ 5m — 6m?. 


3a42x3 — Sax? + 40723, 3a5x? - llatz? + 6a%x, 


Satz? + 16a%23 — 120223, 


118 ALGEBRA. [CHAP. 


147, The highest common factor should always be determined 
by inspection when possible, but it will sometimes happen that 
expressions cannot be readily resolved into factors. To find 
the highest common factor in such cases, we adopt a method 
analogous to that used in Arithmetic for finding the greatest 
common measure of two or more numbers. 


148, We shall now work out examples illustrative of the 
aleebraical process of finding the highest common factor; for 
the proof of the rules the reader may consult the Hlementary 
Algebra, Arts. 102, 108. We may here conveniently enunciate 
two principles, which the student should bear in mind in reading 
the examples which follow. 

I. Jf an expression contain a certain factor, any multiple of 
the acpression is divisible by that factor. 

Il. If two expressions have a common factor, ct will divide 
their sum and their difference ; and also the sum and the difference 
of any multiples of them. 


Example. Yind the highest common factor of 
423 — 3a?- 244-9 and 8x3 — 2x? - 53x — 39. 








x | 4a? — 8a? — 24a -—9 8a? — 2a? — 53a — 39 | 2 
4a3 -- 5a? — 21a 823 — 62? — 48a -—18 
Pee Na 272 — 32-9 2 Gaeta OA Mal? 
Qn — 6x 47 —. 62-18 
3 32-9 x- 3 
32-9 





Therefore the H.C.F. is x—-3. 


Explanation. First arrange the given expressions according to 
descending or ascending powers of x. The expressions so arranged 
having their first terms of the same order, we take for divisor that 
whose highest power has the smaller coefficient. Arrange the work 
in parallel columns as above. When the first remainder 42? — 5a — 21 
is made the divisor we put the quotient x to the left of the dividend. 
Again, when the second remainder 2a?-—3x-—9 is in turn made the 
divisor, the quotient 2 is placed to the right; and so am. As in 
Arithmetic, the last divisor x-3 is the highest common factor 
required. 


149, This method is only useful to determine the compound 
factor of the highest common factor. Simple factors of the 
given expressions must be first removed from them, and the 
highest common factor of these, if any, must be observed and 
multiplied into the compound factor given by the rule. 


XVIII. ] HIGHEST COMMON FACTOR. 119 


Hxample. Find the highest common factor of 
2424 — 2a3 — 60a? - 32a” and 1824 — 62? — 39x? — 182. 
We have 2424 — 223 — 602? — 32a = 2a (122° — x? — 30x — 16), 
and 1824 — 6x3 — 39x? — 18x = 32 (623 — 2x? — 13x - 6). 
Also 2x and 3x have the common factor x. Removing the simple 


factors 2a and 3x, and reserving their common factor x, we continue 


as in Art. 148. 











2% | 6a? — 2a? —- 1382-6 1227 -— x?-30x-16|2 
622 —8x?-— 8x 122° — 4a? — 26x — 12 
2 6a? -— 52-6 3u7— 4r— 4) 2 
62?— 8x-—8 3x7 + Qe 
38x24+2 —- 6x%- 4/-2 
— 6a- 4 


Therefore the H.C. F. is 7(8%+2). 


150, So far the process of Arithmetic has been found exactly 
applicable to the algebraical expressions we have considered. 
But in many cases certain modifications of the arithmetical 
method will be found necessary. These will be more clearly 
understood if it is remembered that, at every stage of the work, 
the remainder must contain as a factor of itself the highest 
common factor we are seeking. [See Art. 148, L. & IL] 


Example 1. Find the highest common factor of 
3x? — 1327+ 23% -21 and 62° +2? - 447421, 
3x3 — 13274+23x2-21| 623+ x2?-44r421]2 
6x? — 26x? + 46x — 42 
27x? — 902 + 63 





Here on making 27z?-90x+63 a divisor, we find that it is not 
contained in 3x3 — 132?+23x%-21 with an integral quotient. But 
noticing that 27x? — 90x + 63 may be written in the form 9(3z?-10x+7), 
and also bearing in mind that every remainder in the course of the 
work contains the H.C. F., we conclude that the H. C.F. we are 
seeking is contained in 9(3x7-10x+7). But the two original expres- 
sions have no simple factors, therefore their H. C. F. can have none. 
We may therefore reject the factor 9 and go on with divisor 
32? - 10x+4+-7. 


120 ALGEBRA. (CHAP. 


Resuming the work, we have 
x | 323 — 13224 234-21 3x2-10e+7 |x 


32° -—10a?+ Tx Ste | ae 
-] — 32°+162-21 —- 3x4+7|-1 
— 327+10a- 7 — 3x+7 
2)6x-14 
Auaias fi 


Therefore the highest common factor is 3x —7. 


The factor 2 has been removed on the same grounds as the factor 9 
above. 


151, Sometimes the process is more convenient when the 
expressions are arranged in ascending powers. 


Hxample. Find the highest common factor of 


Oo 40160? = 0gr oo ee ee (1), 
and AiG LOG Sa" 2. rete ere ee Oo} 


As the expressions stand we cannot begin to divide one by the 
other without using a fractional quotient. The difficulty may be 
obviated by introducing a suitable factor, just as in the last case we 
found it useful to remove a factor when we could no longer proceed 
with the division in the ordinary way. The given expressions have 
no common simple factor, hence their H.C.F. cannot be affected if 
we multiply either of them by any simple factor. 


Multiply (1) by 4 and use (2) as a divisor : 
4— Ja- 19a?- 8a? 12 - 16a — 64a? - 36a? | 3 














5 12 — 21a — 57a? — 24a 
4|20-35a— 95a2— 40a a|5a— Ta? — 12a" 
20-28a-— 48a? 5-— Ja —12a2|5 
— Ja- 47a*- 40a3 5+ 5a 
at 12a —12a?|- 124 
7a ~ 35a-+ 2350? +200a° —12a -—12a? 


3da— 49a2-— 84a? 
2840? | 284a? + 284a3 
l+a 
Therefore the H.C.F. is 1+a. 


XVIII. ] HIGHEST COMMON FACTOR. 121 


After the first division the factor a is removed as explained in 
Art. 150; then the factor 5 is introduced because the first term of 
4—7Ta-—19a?—8a? is not divisible by the first term of 5—7a-— 12a? 
At the next stage a factor —5 is introduced, and finally the factor 
284a? is removed. 


152, From the last two examples it appears that we may 
multiply or divide either of the given expressions, or any of the 
remainders which occur in the course of the work, by any factor 
which does not divide both of the given expressions. 


EXAMPLES XVIII. b. 


Find the highest common factor of 

243 + 322+ a+6, 22+ 4242443. 

2y? — 9y?+9y-—7, y> — 5y? + 5y —4. 

2a? + 8a? — 5a -—20, 6a? -4a?-1524+10. 

a’? +3a2—-16a4+12, a®+a?-10a+8. 

62° — a? — 7a —-2, 2a? —Ta?+x2+6. 

g—3q+2, g@-59q?+7q-3. 

at+a®—20?+a-3, 5a°+3a?-17a+6. 

3y4 — 3y® — 15y?-9y, 4y° — 16y4 — 44y? — 24y/?. 

1524 — 152? + 1022-102, 302° + 12024 + 2027 + 8022, 
2mt+7m?+10m?2+ 35m, 4m*+ 14m? — 4m? -— 6m 4 28, 
3act — 9a? + 12a?-12%, 6a? -6x7-15%+6. 

2a° —4a4*- 6a, a’ +a*—- 3a - 3a? 

x? +442 -—922—-15, x2? —2Qlx — 36. 

14, 9at+2a72?+a4, 3a*- 8a%x + 5a7a? — 2a’. 

15, 2-3a+5a?-2a3, 2-5a+8a?~ 3a’. 


16, 3a?-52°-15a4- 40°, 6a — 7x? -— 2923 — 1224, 


- 


DOONIOARwWNH EH 


ee 
oONnNrH © 


[For additional examples see Hlementary Algebra. | 


GHA PPE Rex 
FRACTIONS. 


153, Tue principles explained in Chapter xvi11. may now 
be applied to the reduction and simplification of fractions. 


Reduction to Lowest Terms. 


154, Rule. The value of a fraction is not altered if we multi- 
ply or divide the numerator and denominator by the same quantity. 

An algebraical fraction may therefore be reduced to an equi- 
valent fraction by dividing numerator and denominator by any 
common factor; if this factor be the highest common factor, 
the resulting fraction is said to be in its lowest terms. 


9407 ¢7a2 
18a°2? — 120223 





Hxample 1. Reduce to lowest terms 
~ 2Aaen* 
6a7x7(3a — 22) 
=. ac 

3a — 2a 


The expression = 





6x? — Say 
Jay — 12 
On a5 A 
The expression = aA _ 2% 
by(3u—4y) 3y 
Note. The beginner should be careful not to begin cancelling 
until he has expressed both numerator and denominator in the most 
convenient form, by resolution into factors where necessary. 





Hxample 2. Reduce to lowest terms 


EXAMPLES XIX, a. 


Reduce to lowest terms : 


a 9 ae —2a_ 9 3ab +b? 
622 —32y * 4a? — 8a? *  6a*b? + 2ab 


1. 


CHAP. XIX. ] FRACTIONS. hes 


Reduce to lowest terms : 























0 eae Baryz Fetes ox? —y" 6 222y7 — 8 
* Bay + 10x72 *  62°y — Qary? * Bay + 6x 
i, pies Gen Gilin OaC eens et = 30 
pa ie 12. * Bea? — 10c2a + 5c" * Bat + 30a” 
2a +b)? a? + 63 2c? + Bcd — 3d? 
Pe el ee ee 
4a° — ab? a? — ab — 2b? c7 + 6cd + 9d? 
x? —4¢ —21 a — I —15 92 +4 -3 
Re ag pe ee pe es 
3274+10c+3 327 — 122-15 227+ 1lla#+12 
3a? — 24 4a? —"Bary? 18a? + 6a2x + 2aa? 
iy | ere Ab et See ele ae ee es 
4a” + 4a —24 227 4+- ay — 1oy" 2 Oe 


155. When the factors of the numerator and denominator 
cannot be determined by inspection, the fraction may be reduced 
to its lowest terms by dividing both numerator and denomi- 
nator by the highest common factor, which may be found by 
the rules given in Chap. XVIII. 


3a? — 1327+ 23% — 21 
15a? — 3827 — 2x%+21 





Example. Reduce to lowest terms 


The u.c.¥F. of numerator and denominator is 3a — 7. 

Dividing numerator and denominator by 32-7, we obtain as 
respective quotients 2? -2x+3 and 5x7-x-38. 
oa? —l3at+23e—21 (82 —7\(x2-2e+3) 272-2243 
15a? — 3827-22+21 (8%-—7)(5z*?7-x2-3) 522-2 -3 


Thus 





156, If either numerator or denominator can readily be 
resolved into factors we may use the following method. 


x? + 3x7 — 4x 
72 — 1827+ 62+5 


The numerator = 2(x?+ 3x - 4) =a(x%+4)(x-1). 


EHxample. Reduce to lowest terms 





Of these factors the only one which can be a common divisor is 
zx-—1. Hence, arranging the denominator so as to shew z-1 asa 
factor, 

a(a+4)(a—-1) 
Fax —1)—1lx(z—1)—d5(x—-1) 


ee ee le ale ts) 
~ (@—-1)(7a? — lla -—5)” 7a2?-lla—5 





the fraction = 





124 ALGEBRA. [coHaP. 


EXAMPLES XIX. b. 


Reduce to lowest terms : 




















1 x? — 4? +24 —2 9 oe+at+2 
© “Bat4 72242 * @—402+5a—-6 
9 yey ly 3) 4 m>—m?—2Qm 
* 3y2+4y?+4y+1 * m—m?-—m—-2 
5 a? — Qab?+ 21 b3 ; 6 Qa? — aa = 90° ‘ 
* a? — 4a7b —2lab? * 323 — 1l0ax? — Ja2x — 4a? 
7 5a? —4a-—] : 9 +27 — 12cd? = 9d? : 
* 923 — 32741 * Qc? + 6c2d — 28cd? — 24d3 
9 aA —21e+8_ 10 ap Oy ly = oy". 
* 8a4—21a? +1 * y+ Ty’ + 3y?- lly 
ll Rea ings 12 2 —5ax — 4a? + 323 





2-a +923 4+ 4a + 922+ 423 — Bat 
[For additional examples see Hlementary Algebra. ] 


Multiplication and Division of Fractions. 


157, Rule. To multiply together two or more fractions: 
multiply the numerators for a new numerator, and the denome- 
nators for a new denominator. 


Ce Caae 
Thus B* dba 
Sunilarly, as del sl 


BaF bdf’ 
and so for any number of fractions. 
In practice the application of this rule is modified by re- 


moving in the course of the work factors which are common 
to numerator and denominator. 


2a? + 3a y 4a? —6a 
dae) 18a 15) 
a(2a +3) 2a(2a — 3) 
4a 6(2a + 3) 
_2a-3 
oom 


by cancelling those factors which are common to both numerator and 
denominator. 





EHxample. Simplify 





The expression = 


158, Rule. Zo divide one fraction by another: invert the 
divisor, and proceed as in multiplication. 


XIX.] FRACTIONS. 125 











d ad 
Thus sa a oe ah 
by di bd -e* 06 
Fe Vy PC) ae 
Example. Simplify Cea aed pae A ASS aD er 
ax — az 9u7-4a? 38ax+2a? 
: 6x? — ax — 2a? x-a 3ax + 2a? 
Th Ss = : x ) 
e expression en ara eo a 
: _ (8% ~2a)(2a +a) u-% ee C(O -+ 2a) 
. a(x — a) (84+2a)\(8a2-2a) Weta 
=, 


since all the factors cancel each other. 


EXAMPLES XIX. c. 









































Simplify 
7 zea pe, 9. ee 2b — Bab® 
te ARG UG e724 2-12ab a*b?-4 
9 2c? + 3cd , e+d 4 5y - lOyees ] -Qy 
* 4c? - 9d? ° 2cd - 3d? * 12y? + 6y? © Qy + y? 
5 x? —4 Nate 6 b2 uy oes 16a? 
* 9244744 °> 424+2 1 3b=—da~ = $2295 ~ 
7 Eee a 8 aU heey ey = 24 
*  @?+5e+4 a74+3%+4+2 ; pees y+ 6y+9° 
G74 2) ee 40 = 2) a*—3a—2_ 3a?-8a-3 
Drees gyee gr eaee 10, x pee 
Ay 6? + 125 . 2O0 tn 12 38m?—m—-2 , 4m?+m—-6 
* 5b?+24b-5 63—5b2+25d *  3m2?+8m+4° m+2 
13 merely Os 5p +6 ee 2p-- 15 
“ “p=9 p?-5p p-4 
64a72—=1 4#7=49 . -w=7 
14, ie eat Ee ee Sa 
15 407+ 44-15 x+8 ; 2a? + 5x 


Piet -46 (abe eis 6 
16 a? + 8ab — 9b? a? — Tab +12? a+ ab fab? 
, ard 27b ai? — b? a? — 3ab — 4b? 
17 — 16a? 4 x*+ae—2007 , x? - pee Cat 
; ae —ax—3002 ax2+ 9a2x +2003 x? + 8ax + l5a2 
(a—b)?-c? a?+ab+ac . (a+b)?-¢? 
18, Gee ele (a—c)?- 0? 7 (a+b+c)? 














[For additional examples see Llementary Algebra. ] 


CHAPTER XX. 


LOWEST CoMMON MULTIPLE. 


159, Derinition. The lowest common multiple of two 
or more algebraical expressions is the expression of lowest di- 
mensions which is divisible by each of them without remainder. 


The lowest common multiple of compound expressions which 
are given as the product of factors, or which can be easily 
resolved into factors, can be readily found by inspection. 


Example 1. The lowest common multiple of 6x°(a—2)?, 8a®(a — x), 
and l2ax(a— 2x) is 24a%x?(a — x)°. 
Hor it consists of the product of 
(1) the L.C.M. of the numerical coefficients ; 


(2) the lowest power of each factor which is divisible by every 
power of that factor occurring in the given expressions. 


Haample 2. Find the lowest common multiple of 
38a7+9ab, 2a*- 18ab?, a® + 6a2b + 9ab?. 
Resolving each expression into its factors, we have 
3a? + 9ab = 8a(a+3d), 
2a? — 18ab? = 2a(a+3b)(a — 3d), 
a? + 6a*b + Yab? = a(a + 3b)(a+4+ 3b) 
= a(a + 3b). 
Therefore the L.C.M. is 6a(a+3b)?(a— 3b). 


Example 3. Find the lowest common multiple of 
(y2—xyzy, ya — x3), x44 Qad-+ a2, 
Resolving each ee into its factors, we have 
(yo —ay2)= {yee 2) =P 2-2), 
yy? (202 — 203) = y?ar( 2? — x?) = wy?(z — x)(z+2), 
ZA 4 Dare? 4 9022? = 2( 2? + Daz + 2?) = 22(z-+-00), 
Therefore the L.C..M. is wy?2*(z+a)?(z-—a)%, 


CHAP. XX.] LOWEST COMMON MULTIPLE. 127 
EXAMPLES XX, a. 
Find the lowest common multiple of 
1, a’, a-a’. 9, x7, 2? — 323, 3. 4m?, 6m —8m?2. 
4, 627, 2*+327. 5, b+), b-b. 6, «2-4, 2°+8. 
I, 9a?b-b, 6a? +2a. 8, B-k+1, B-1. 
YY. m?-5m+6, m?+5m-14. 10, y?+8y?, y?-9y’. 
ll. 2?-9x2+14, x?+4e-12, 12, 2+27y*, x?+xy- by’. 
13, 067+9b+20, b?+6- 20. 14, c?-3cx-182?, c?-8ex+ 122%. 


15, a*?-4a-5, a?-8a4+15, a?- 2a?- 3a, 
16. 2a?-4ay-16y?, 2?-6ary+8y?, 3x? - 127°. 


17, 3a3-12a?x, 4x7+1l6arv+16a7, 18, a’c—ac®, (a2c¢+ac?)*. 


19, (a%x—2ax*)?, (Qax — 42)2, 90, (2a-a7)3, 4a?-4a?+a4, 


91, 2u?-x-3, (2x -38)?, 4z7-9. 

99, 2x2-Tx-4, 6x?-Tx-5, x2? - 8274+ 16x. 

28, 10x°y%(x?-y*), ldy*(a—y)?, l2a%y(a — y)(a? — y?), 
94, 2a? +x-6, 7x?+1lxa-6, (7x7- 32). 


95, 6a? — Tax -3ax", 10a?x - llax?- 6x, 10a? -2lax — 102". 


160, When the given expressions are such that their factors 
cannot be determined by inspection, they must be resolved by 


finding the highest common factor. 


Hxample. Find the lowest common multiple of 
2a + a? — 20a? — 7x +24 and 2a4+ 3x° — 138x2-7a+15. 


The highest common factor is z?+2z.- 3. 
By division, we obtain 


Qa + a3 — 20a? — Ta 4+ 24 = (x? + 2a — 3) (Qa? — Ba - 8). 


Qart + Ba3 — 13x? — Ta +15 = (2? + Qa — 3)(22? — w — 5), 


Therefore the L.C.M. is (xz? + 2a — 3)(2a? - 3x” —8)(2u?- a” -5). 


128 ALGEBRA. [CHAP, Xx. 


EXAMPLES XX. b. 


Find the lowest common multiple of 


], «-Qa?-132-10 and x - a? 10x- 8. 

2, y'+3y?-- 38y-9 and y*®+ 3y? - 8y— 24. 

8, m+3m?—-m—3 and m?+6m*+1l1m+6. 

4, 2Qat-2e3+274+3x2-6 and 4a4-2a°+ 3x -9, 


. Find the highest common factor and the lowest common 
multiple of (%—«?)®, (~?-a°)?, a? — a4 
6, Find the lowest common multiple of (a*- ax), (a?+az)’, 
(aa — x7). 
7, Find the highest common factor and lowest common multiple 
of Ga?+5¢-6 and 6a?+x%-12; and show that the product of the 


H.C.F. and L.C.M. is equal to the product of the two given expres- 
sions. 


8. Find the highest common factor and the lowest common 
multiple of a?+5ab+6b*, a*—4b?, a? — 3ab?+2b°, 


9, Find the lowest common multiple of 1-a?-a4+2° and 
1422 +2? — at — 2, 


10. Find the highest common factor of (a? —4ab?)?, (a?+2a°b)°, 
(a*x + 2abx)?. 


11, Find the highest common factor and the lowest common 
multiple of (3a? —2ax)?, 2a7xa(9a? — 4a"), 6a? — 18a7x? + bax’. 


12, Find the lowest common mrltiple of #+a°y+ay?, wy -y', . 
amy + ay? + xy, 


[For additional examples see Hiementary Algebra. ] 


CHAPTER XXI. 


ADDITION AND SUBTRACTION OF FRACTIONS. 


161, To find the algebraical sum of a number of fractions 
we must, as in Arithmetic, first reduce them to a common 
denominator. For this purpose it is usually most convenient to 
take the dowes¢ common denominator. 


Rule. Zo reduce fractions to their lowest common denomin- 
ator: find the L.C.M. of the given denominators, and take it 
for the common denominator ; divide it by the denominator of the 
first fraction, and multiply the numerator of this fraction by the 
quotient so obtained; and do the same with all the other given 
fractions. 


Hxample. Express with lowest common denominator 
2a(x — a) 32e(a2 — a?) 
The lowest common denominator is Gax(x%-a)(w+a). 
We must therefore multiply the numerators by 32(v+a) and 2a 
respectively. 
Hence the equivalent fractions are 
15x7(a~+ a) er 8a? 
6ax(«-a)\(x+a) 6ax(x—-a)(v+a) 





162, We may now enunciate the rule for the addition or 
subtraction of fractions. 


Rule. Zo add or subtract fractions: reduce them to the 
lowest common denominator; find the algebraical sum of the 
numerators, and retain the conmmon denominator, 


1 a,c _ad+be 
zou ue he 
and a_¢_ad—be 

6d bd 


130 ALGEBRA. (citar. 


168. We begin with examples in further illustration of 
those already discussed in Chapter x11. 
: 2e+a 5a?-4ax 
Hxample 1. Find the val fee ae 
xample ind the value of + 2-5 
The lowest common denominator is 9a. 


Therefore the expression = da(2x + a) + Sat — dam 


9a? 
= 6Gax + 3a?2+4+ 52? -4ax tt 3a? + 2ax + 5a? 
i 9a? > 9a* 


x—-2y 3y-a 34-240 
ee eR ae 
The lowest common denominator is axy. 
Thus the expression = COA at eae 
axy 
ac —2ay + 3xy -ax—38xy + 2ay 
= waxy 
= 0, 
since the terms in the numerator destroy each other. 


Hxample 2. Find the value of 





Note. To ensure accuracy the beginner is recommended to use 
brackets as in the first line of work above. 


EXAMPLES XXTI. a. 
Tind the value of 
Be-1 2+3 2e-1 



































1 a-2 a-l at+5 9 
Sure oe eOe lpia GEE ees 
peor iy LO MueuNaG caret. 
5 eg! 2-2 6 22-5 2-4 a2 —dAg 
we ye yp RS) Mt Mel eu: ener 
Ye ee atx at2ex x-5a. 
1. ye | wet xy the 2a 3a. | 64 
i 3 2 joe a PAs 
9. 2a°—-5a a pee reads 10 See. Ey eee, 4x? 
a a" a? y oe Bry 


R20 30 aC, ab-be a 2a®-ab 
TSE Re Ope a Se tea) 

13. ay-axy+4e_ 1 _ a. 14, Sa era 

Quy Ee a" be ca 














XXI.] ADDITION AND SUBTRACTION OF FRACTIONS. Tal 


164, We shall now consider the addition and subtraction 
of fractions whose denominators are compound expressions. 
The lowest common multiple of the denominators should always 
be written down by inspection when possible. 


22-30 2x-a 


Hxample 1. Simplify 5 : 
x-20 x-a 








The lowest common denominator is (% —2a)(z—- a). 


Hence, multiplying the numerators by x-a@ and x-2a respec- 
tively, we have 


the expression = es ee) 








(2 — 2a)(x — a) 
_ 2a? — 5am + 8a? — (2a — Sax + 2a?) 
he (a —2a)(%— a) 


Qu? — Fax + 3a? -— 27+ 5ax - 2a? 


(x —2a)(x—-a@) 





a 
> (a —2a)(x — a) 





Note. In finding the value of such an expression as 
— (2% -—a)(x - 2a), 


the beginner should first express the product in brackets, and then 
remove the brackets, as we have done. After a little practice he 
will be able to take both steps together. 


Example 2, Find the value of ae + ame 





The lowest common denominator is (a -- 4)(~% +4). 
(82% +2)(x+4) +(x —5)(x - 4) 
(7 -4)(~+ 4)? 

_ 3x7 +14e+8+2?—9x+20 
(x —4)(~ +4)? 
_ 447+ 5a +28 
(x—4)(a +4)? 


Hence the expression = 








165, If a fraction is not in its lowest terms, it should be 
simplified before it is combined with other fractions. 












































132 ALGEBRA. [cHaP. 
B be Wa are v?+5ey—4y?_ xy — 3y? 
cample. Simplify 2 16,2 a wrap, 
2 — Ay? = 
The expression =~ ey. ay = ED 
a*—l6y? (w+ 4y)(a — 3y) 
Ua Oly tee oe 
x” — 167? x+4y 
+ bay Ay yee ay) 
x? — 16y? 
_v+5ay —4y—- xyt+4y? 
x? — 16y 
_e+4ay  wetd4y) & 
2 —l6y? (a+4y)(a—4y) x—4y 
EXAMPLES XXI, b. 
Find the value of 
] 1 1 1 i 1 
1. a—2*a-3 2. asd ae a 3. b-2 642 
a b Gt be a+3 a-3 
4, x-a «x-—b 5, ate 'a-@ 6. a-3 a+3 
z he 3a 1 <a x — Qy 
1 e-1, 2?-1 8 a—-4° a+2 9. a2 — dy! e+ 2y" 
] ] 3a 2a 
10. a(a—b)* a(a+b) 1, Qu(a—a) 3a(x+a) 
5 Aa 1 3(y +2) 
12, 2—-2° (%—2)(@+1) 18, y—2y-3 y—y-6 
] a 3 2a b 2 
14, tou a’ d—a =a) Lo; z+y “rye 16, (b+ 1 1p AS | 
2at+y 2e-y b+e b —2¢ 
Me ay (wry 18 Brera BE 
x xy a*+ 2a a 
19. xy—y*? w—a*y 20. a+a—-2 atl 
4a7- bh? 4a oe ] 
21, Qab—b2 2a+b 22, +1 +1 
2b-4 1 LID? e442 — Dyt 
23, B+ b+2 a4. Prayt+y* wy-y* 




















XXI.} ADDITION AND SUBTRACTION: OF FRACTIONS. 133 


Find the value of 


] iL ] il 
20. a2—6a4+9 a?—5a+6 26. a? —3a+2" 08 +3a—10 





—2 a—-6 ] x? ] 
Ja tate, Ps teat 5) i 
OT, w+2-Z 9g, 4487 Pog, 86 99, 4] 
] 2. ] 3 ] De 
80. Re Ot ed 31, Gow a 2 aLG. 


166, The following examples furnish additional practice in 
the simplification of fractions. 





: ae ee 4 2 ,7b+5 
v0) le. lif = , 
xample. Simplify 3543 cat pay 

ie 
The expression = Ae ee Rh 





3(6+1) 5(b-1) 02-1 

_ 20(6-1) — 6(6+1) +7645 

» 15(b2 - 1) 

_ 210-21 _21(b-1)_ 7 
15(62—1) 15(b?-1) 5(64+1) 








167. Sometimes the work will be simplified by combining 
two of the fractions together, instead of finding the lowest 
common multiple of all the denominators at once. 


3 < ] ye Gers 
S(a—x) S(a+x) 4(a2+22) 





Hxample, Simplify 


Taking the first two fractions together, 


the expression = Aes et 2 oe zs ae eu 
8(a* — x?) 4(a* + x?) 





(Reeve teh en 


4(a*-—a?) 4(a?+ 2?) 


_(&+2%)(a? + x?) — (a — 2x)(a?—2") 











4(a* — a*) 
_ a + 2a%ax + ax? + Qa? — (a? — Qa22 — ax? + 2x3) 
4(a* — x*) 


_ 407x%+4+2ax?  ax(2a+2) 
4(a4—at) _ Qat—x4)- 





134 ALGEBRA. [CHAP. 
EXAMPLES XXTI, c, 
lind the value of 
ee oe 8 ee 8n i ee a a 
MM Oee31 8a 1 Sax? 1 " Qa+3e Qa—3c 4a?-9c* 
1+2a 3a7+2a Qa 4 4a° — 9a 
= oer Re es ; 
3. 3 — 3a 790 4. 9-6x 6442 27-122? 
5 1 2a a? G 2a fh 2 
* g-a (#-a)? (2-2) * (a+1? (@+1)4 (a+1) 
a a ab 1 1 1 
1 (a—b}2 a-b (a—b)/ 8. Sam any, Set 
fe og ei ee 
° Qp—y-38 Ww+y-1 ' 4432-22 3442427 
1 1 2 3 x 1 
° — + — Soe Be ae ® jeS 3 ° 
iT ge=1) @ez+1) 21 12 (o—-2) 2-4 eZ 
D4 3 1 
18, 3-a (24a)(3—a)(1+2a) (3 ~a)(1+2a)" 
14, Yao = oAe Ss) ae yaa : 
(Y—3)ly¥—4) (y—2)ly—4)e oy =2)\y = 9) 
15 LoS es Oe ee eed ee 
’ T-x2% (l-x)\(2-a) (1-x\(2=—2)(84+2) 
2 3 1 
16, x? — Bay + by? fae Cee 
17 5a OA: 3 yea OAL 
* 6(a?-1) 2(a?4+2a-3) 30?46a+3 
18 x—-5 20. eo. 
* ode —5 w24+2e x24+2-6 
b2 34. p38 
10: =, 2 2 oa a 
a-b a®?+ab+b? a—b 
Oe ee Dame ee 
* 98427) 2? -38249 243 
PALA en ce eee as 
* (x-y) 2 +2ay+y? xt-22°y*+y4 
x a ax 
HES vas tap (ae 
J 1 3 a x 3 
es ‘ : — Es : 
BO to ee 8 24 4(l+a) 4(1—a)  2(1+2°) 













































































XXI.]| ADDITION AND SUBTRACTION OF FRACTIONS. 135 

















25. Oe Tec OTS 26. eben) Poa 
a. ba Pr 28. ar et ee 
0. ee i a 


168. To find a meaning for the fraction = we define it as 


the quotient resulting from the division of —a by —b; and this 
is obtained by dividing a by 6, and, by the rule of signs, pre- 
fixing +. 





Therefore TEES, si Re aah aE i) 
B= +5535 (1) 
Again, —" is the quotient resulting from the division of —a 


by 6; and this is obtained by dividing a by }, and, by the rule 
of signs, prefixing —. 


Therefore BLISS, So Ae, Mile ttn thy Re 2). 
D F (2) 
Likewise = is the quotient resulting from the division of 


aby —6; and this is obtained by dividing a by 0, and, by the 
rule of signs, prefixing —. 


Therefore 


These results may be enunciated as follows : 

(1) Jf the signs of both numerator and denominator of a 
fraction be changed, the sign of the whole fraction will be wn- 
changed. 

(2) If the sign of the numerator alone be changed, the sign of 
the whole fraction will be changed. 

(3) If the sign of the denominator alone be changed, the sign 
of the whole fraction will be changed. 


UE Came Ui) Oi ea — Ge 


Example l. fe - eee oe 
Z Ye Yan e) =Yrr Ly 





x — x —x¢+a? xu -— 2 


Die a Di Qy 





Hxample 2. 








136 ALGEBRA. [CHAP, 


an 
=A at” 7 oa 4 


Example 4. Simplify —“— + aE abe as o) 
eta xex-a ar-az 


we) 
5 











Haample 3. 





Here it is evident that the lowest common denominator of the first 
two fractions is x2?- a’, therefore it will be convenient to alter the 
sign of the denominator in the third fraction. 





: a 22° a(8e—a 
Thus the expression = ——_ i ( = 2) 
Le Ome a ae Oe 


_a(@—a)+2x(x +a) — a(8a—a) 








= By 5) 
x? -~ a? 
ax — a2 4+ 2x? + Qax — 8ax +a? 
= v2 — ae : 
Qa? 
~ 2 g 


1 1 1 


Example 5. Simplify Coes -+ ECE + Caos 





Here in finding the L.C.M. of the denominators it must be 
observed that there are not sia different compound factors to be 
considered ; for three of them differ from the other three only in 
sign. 

Thus (a—c)=—(c—a), 

(b—a)=— (a— Dd), 
(c—6)=-—(b-c). 
Hence, replacing the second factor in each denominator by its equi- 
valent, we may write the expression in the form 
=~ 1 _ ] R) 1 
(a-b)(c—a) (b-c)(a—b) (c—a)\(b-c) 

Now the L.C.M. is (b-—c)(e—a)(a—d) ; 

—(b-—c)-(c-a)-(a—b) 
(b—c)(c-a)(a—b) 

ial ee eat nee) 
(b-—c)(c- a)(a—b) 

20) 





and the expression = 








Note. In cxamples of this kind it will be found convenient to 
arrange the expressions cyclically, that is, so that ais followed by b, 
b by c, and ¢ by a. 


XxI.] ADDITION AND SUBTRACTION OF FRACTIONS. 137 


169, If the sign of each of two factors in a product is changed, 
the sign of the product is unaltered ; thus 


(a—2x)(b-#)={-@— a) ~ @—b)}=(#@—a)(w— 6). 
Similarly, (a—x)?*=(x#—-a)?, 
In other words, in the simplification of fractions we may 


change the sign of each of two factors in a denominator without 
altering the sign of the fraction ; thus 


igureaertr 
(b—a\(e—b) (a—b)(b—c) 


170, The arrangement adopted in the following example is 
worthy of notice. 





1 1 Qu 4x8 
Example. Simplify RRs PACE RW ERard tae BUNTY 





Here it should be evident that the first two denominators give 
L.C.M. a?—2?, which readily combines with a?+2? to give L.C.M. 
a*—a4, which again combines with at+a* to give L.C.M. a’-2%, 
Hence it will be convenient to proceed as follows : 


a+x—-(4-x)_ 











PENG EXpressiOn = ——— a es ent k 
a*— 2 
we EE 
nea ae 
doy Ta Aa 
at—xt at+at 
Sar 


EXAMPLES XXI. d. 


Find the value of 














5 32 4 — 13a D) 10 ee he il 
1422 1-2x 4227-1 POS G75 30, a= 3: 
5a 1 1 2y 5 12y+8 
: i ; + 5¢ 
3. Giaialh Mla): Staal} 4 Qy — aa 6y +9 27 — 12y" 
5 +a x-a 4ax 6 8—2c , 2c4+3 12 








x-a “+a a2? 34+2c 2-3 4c?-9 


138 ALGEBRA. [CHAP. XXI. 


Find the value of 














Tap eeipiiee 8 aoptaset roe 

i ee 10, et 

nt CEE CSAS CET MATER CEE) 

12. pene ecg Woz SS 

13. atacer eat eo Ce ae 
ae: Ces eo 16. Gaylene) Ya aly =a 
ey are) ay tesa aes 

1. tt 





ate a-% x-a Wwta 


20 eee ee 
ate 3x+a “w-a a-—3ex 























Ney earner Berea EN OE 
22 en cee 
aa canes (q ait —D =a a) 
26, oT +2) Fare a) z Had a?) : = 7 
8, oto teoar at 


g-27°Q+ap" Qa a+2 


CHAPTER XXII. 


MISCELLANEOUS FRACTIONS. 


171, Derinition. A fraction whose numerator and deno- 
minator are whole numbers is called a Simple Fraction. 

A fraction of which the numerator or denominator is itself a 
fraction is called a Complex Fraction, 


a 
Thus . are Complex Fractions. 
d 


In the last of these types the outside quantities, a@ and d, 
are sometimes referred to as the extremes, while the two middle 
quantities, 6 and ¢, are called the means. 

Instead of using the horizontal line to separate numerator 
and denominator, it is sometimes convenient to write complex 
fractions in the forms 


Simplification of Complex Fractions. 


172, It is proved in the Elementary Algebra, Art. 141, that 





a 
ba EOy Ga. 00, 
inl ae Done be. 
d 


The student should notice the following particular cases, and 
should be able to write down the results readily. 


Ley OL 8 
a b Gad 
b 

a, 1 ; 
ree p70 b=ab 
b 


140 r. ALGEBRA. [cHap. 


173. The following examples illustrate the simplification of 
complex fractions. 














2 

e+e 
ELxample 1. Simplify ; 

tine 

2 4 9 2 ens 

The expression (2+ se Z (« > a Leta? ata 

x uw? C ye 

: : _9 
Example 2. Simplify = a 

Hoe 


Here the reduction may be simply effected by multiplying the 
fractions above and below by 6a, which is the L.C.M. of the 
denominators. 


18 + 2a? — 12a 
a?+3a-18 
_2(a?-6a+9) 2(a-3) 
\(a¢6)(a—38) “a+6 


Thus the expression = 

















a2 +0? a? 0? 
Example 3. Simplify ere 
a0 aD 
TI des (Ga Oe ae 4a°b* 
ae numerator (a?+ B5(a2 02) = (2 ED a2 BY 
similarly the denominator = ne 
: 4a?b? dab 
H the fraction = ae 
ence the fraction (®t B\ ew) * @Lbanb 
a 4a7b? (a+ b)(a—b) 
(a@2@+e)\(a2—b)~  4abd 
= 20d 
a2+b? 


Note. To ensure accuracy and neatness, when the numerator and 
denominator are somewhat complicated, the beginner is advised to 
simplify each separately as in the above example, 


XXII. ] MISCELLANEOUS FRACTIONS. 141 


174, In the case of Continued Fractions we begin from 
the lowest fraction, and simplify step by step. 














Haample. Find the value of : ; 
ae eee 
oe 
The expression ek ae Eten ed inl 
7 3 ceo — x) 
Q2-24+2% 2-2 
Si 
iS 1 1 
8-44-3432 5-a@ 
2-2 2-2 
2+2 
62 


EXAMPLES XXII, a, 
Find the value of 












































1 a 
1, pry Us oa o - is 4, —. 
+2 b= = ee | = 
e Zz d a> l-a 
aes en pel P_y 
ae te NOE d 4g 
5, ea 6. y a le a 1 8. ab Tr 
ie OL xy Bard Pa p 
a+? -5 y-34+2 aoe = 
9, 10. . ak, ea 
ee eee noe 
a2 a y n 
oan p-2-—8, ee 
12, ———. Bo ——— ss 14, — 
x-44_l% Dens BF coat 
243 b+3 at abt R 
c+d c-d qa - 222 2+3 «+3 
ae TG l-ab a 4 
15. c+ad c-d 16. 1 a(a—b) 17. x-3 2-3 
c-d'ct+d L—ab z—1 
1 1 eek 
18, 1 * tomes 19. C+ ¥ 20, a c 
1+- x — 4--= 
a a ad 


142 ALGEBRA [CHAP. 


Find the value of 























] 
21, —— 22, ——— 23, ——. 
]-—= L+H lee 
x x y 
Rew 
D4, 2 “ ~ 95. a 26. “ ~ 
es l= ee eae | 
ie loa dae 
Lite, 2-y ip 
22] 8a —2 
i Sera 28, ae 
De ia Si Oe ee 
lee [eae 
x—-1 3a — 2c 


175, Sometimes it is convenient to express a single fraction 
as a group of fractions. 


Day — lay loy? bay 10ry? | be 





Example. 


102: ~ 10x%y?  10x%y* 102%? 
Lael ae, 
ee eae 
2 pe aeaete 


176. Since a fraction represents the quotient of the nume- 
rator by the denominator, we may often express a fraction in an 
equivalent form, partly integral and partly fractional. 


+7 (#+2)+5_ 44. 5 








Hxample | 








x+2 x+2 x+2 
Example 2. 3x-2 _ 3(@ +5) — 15-2 _ 3(@4 +o)— 17 17 ae LZ ate 
2+5 x+5 x+5 x+5 
Example 3. Shew that 2Qu*— Te—1 =22-—1= meal 
she © 2-3 
By actual division, x -3)2x4?-TJa-—1(2xr-1 
200 — 62 
- x-l 
—- £+3 
-4 
Thus the quotient is 22-1, and the remainder ~4, 
Peas cs 
Therefore vast adel Me 2a—-1- Lies 


«“-3 x-3 


XXxII.] MISCELLANEOUS FRACTIONS. 143 


177. If the numerator be of lower dimensions than the 
denominator, we may still perform the division, and express the 
result in a form which is partly integral and partly fractional, 








Example. Prove that j —_ = 2% — 6x? + 182° - tees 
By division 1+322 ) 2e ( 2a — 6a? +1825 
2x + 623 
Fe3 
— 62? — 1825 
1825 
182° + 54x7 


— 54x47 
whence the result follows. 

Here the division may be carried on to any number of terms in 
the quotient, and we can stop at any term we please by taking for 
our remainder the fraction whose numerator is the remainder last 
found, and whose denominator is the divisor, 

Thus, if we carried on the quotient to four terms, we should have 


Qa ; 16229 
g PEs Ont G8 1905 <4 a 
aS ree ch me aia 





The terms in the quotient may be fractional; thus if 2? 
is divided p x*—a', the first four terms of the quotient are 


ied ae ea 
ae sua and the remainder is — 
A ei gw 
CO ae pe 


178, The following exercise contains miscellaneous examples 
which illustrate most of the processes connected with fractions. 


EXAMPLES XXII, b, 
Simplify the following fractions : 











l 1-23 9 12a? +x2—-1 ,1+6%+92x? 
* 14294 2424 ge * 1-824 1622" 1622-1 ~ 
a+b. 4ab a+b 2a 
3. i teas. = at 4. o—ab—26? a?—4o° 
5 a—1 xt+e2+1 6, (w+y)?_(#- (w-y) 





gz-l x+e+) CV-Y “bY 


144 ALGEBRA. [crrap. 


Simplify the following fractions : 









































7 aba" — aca + bay — cy 9 “I ee! )- a 
ax*+xy—-ax—y ‘ g\a-x% a+3x/ a+3u 
2 3 5 
9. 6848 bat alle +08, Oneal: 
Se 02+ a* 1 
3a.) a Cee 
10. 3a+1 “ORE H a? + 23 Sore 
1 1 eee 
=, z—1 
12, ae a 18, 92 _ 3(a? — 2ia+ 54) 
cd(a? + b?) + ab(c? + d?) ( x ee 
1d. cd(a? — b*) + ab(c? — d2) 16. lta a a 
16 ] ., 1 
: —8a22+4a? 23 — aa? — 4074+ 403 
2 
Bae oO 2(422) 
17 2 eae mW 0 ngs 
e = 3 * ° = 
204 a’+8 lig 3 
a a 


19 223+ 42-32 | 5a?-8x-21 , 2x?-3x-9 
* 8542+ 24a -—35° w+ 7a? -S8a ° 722+5la—-40 








Cat Pr Te p= Dad ee 
MS IISA Cae al 1rd 


21 ye ee ee 
; “a-y xt+yl \x-y “t+y 2a7y + 2ry" 


a?—(b-c)?  b?-(c-a)? , 2 -(a—b/y 
22, (ctaf—B atbpie (b+c)?-a 


23, (= +¥)( : )- ye a el 
YING) ae Pay ae x+y 


94. y a1) te a*—4a+3, a? — 4 — tale 


a+a-6 a2 —-4a4+4 weet ae 



































Waste at sea ~4ax 
25. (55 Toa z) (sa saa) 6a*(x - 2a)(% — 3a)’ 


XXII. ] MISCELLANEOUS FRACTIONS. 


Simplify the following fractions : 
































i] ] i 2a? 
20. 6a —6 6a+6 30243 8a4+3° 
97 4ah? Ae 1 a) a x 1 
* 2a4+32b4 8a+16b 4a7+1602 8(2b-a) 
3b7-+6 .2b=7, 26°=3b ; 
A Rae EB at 
28. Geb t 195 28 43" 
99, att 
ND I HI 
x x y y Z 2 
Dga 2 
LT we FES een : 
20 (ger) xy 9] m m+m 
ee “- Pigede ie 1 ns 
x) \a a m4 m 
ab ac ] ] if 
32. (sea) A aie a 
[ba bes ha 
re b Cc 
eli pends 6 
99 Ltac = 1 -—ae Eee a) 
ny | sree cose aa eee) | a 
l+ac l-—ac 
1 ] 
(30-- wee oe 1 
34. ul to 3a Riera iy) TPR) 
a a 


145 


CHAPTER XXIIL 


HARDER EQUATIONS. 


179. Some of the equations in this chapter will serve as a 
useful exercise for revision of the methods already explained ; 
but we also add others presenting more difficulty, the solution 
of which will often be facilitated by some special artifice. 


The following examples worked in full will sufficiently illus- 
trate the most useful methods. 


6x-3 32-2 

2+] e+5° 

Clearing of fractions, we have 
(6a — 3)(v7+5) = (8a —2)(24+7), 
6x7 + 27x — 15 =62?+17x-14; 


Example 1. Solve 





1022 1% 
1 
ea=—.,. 
10 


Note. By asimple reduction many equations can be brought to 
the form in which the above equation is given. When this is the 
case, the necessary simplification is readily completed by multiply- 
ing across or ‘‘ multiplying up,’’ as it is sometimes called. 


87+23 5a+2 2%4+3 
Example 2. Sol ie = eat; 
xanrple olve 50 te hae 1 
Multiplying by 20, we have 
8a +23 _ 20(5% + 2) 











cxaisd Ge = 82x + 12- 20. 
: a 20(52 +2) 
By t t BY fee ee 
y transposition, Nera 
Multiplying across, 93x7+124=20(57+2), 
84= 725 


¢= 12, 


CHAP, XXIII. ] HARDER EQUATIONS. 147 


180. When two or more fractions have the same denomin- 
ator, they should be taken together and simplified. 
24-52 Sx-49 28 
E. Lax kee OLY 6 ane = — 13. 
aample NGS ary tap tame eae 
By transposition, we have 








8a — 28 — (24 — 5x), 
ee eect a ea Ook 
4— : x-2 
Gree dae 








4-¢ 2-2 
Multiplying across, we have 
3a — 5a?-6+102 = 16-424 20x — 52? ; 


that is, — 32 =22; 
Bix _ 22 
= 

Example 2. Solve a EN ea am “2-7 


we 108 wae Boe ESO. 


This equation rent be solved by at once clearing of fractions, but 
the work would be laborious. The solution will be much simplified 
by proceeding as follows. 


The equation may be written in the form 
(10) Pl (e = 6) oe Ge ye (ae 9) 2 
Sanyo © Eee yh een ay 
whence we have 














2 2 2 2 
i el 
Pa (ae ae es Seay, =9t 
A A 1 1 i ] 
h = : ‘ 
al la #210) eEbiao7 2-9 
Transposing ! st Vis Rie a ec ae 
: x-10 x-7 «x-9 x-6 
3 3 


(a —10)(z7—7)° (%—9)(a—6) 


Hence, since the numerators are equal, the denominators must be 
equal ; 


that is, (a —10)(a-—7)=(a%-9)(x%-6), 
x? —17a+70=27-15%+543 
TG es da 


ee “& 8. 


148 


ALGEBRA. 


EXAMPLES XXIII. a. 


Solve the following equations : 


























ape! 9 ie Gabe 
5e-9 42-10 © 62-17 42-13 
7_3-4a 1 ee, 
9 4-52 4, Fhe ees ae 
Bu-8_5x+14 F 8a-1_4a-3 
pad  “aToo °° 6x42 32-1 


220-12 _ 4, 3e+7 9n- 22 38a-5_ 


























8a-5 4248 8. On 5 One] 
8x-19 1_3e-4 10, 2#t2 1, 6e=1, 
42-10 2 2x41 = 3(¢ 21) ee ook 
e-5 22] oe =1e 52 
2 3x42 10 5 
Bu-17 , 2¢-11_28_3e-7 
13 -—4a 14 42 Ai 
4x-—3 1-92 _4%+3 1 
ald eh tee en gOd ea 
eee eae al 
xt+l “+2 %x4+3 3x2+6 
Be aes 2 es 
zx-4 3x-18 47-16 xw-6 


Liye eee ae 
“X+6 3a+12 2%+10 6(@+4) 
2-1 #£-5_ “2-3 0-7 
e-2 x-6 «x-4 “2-8 

1 és ] 1 ] 


oO ee l7. 2 llete— 15: 














§x-64_ 4%-55_2x—-IT x- 


6 
2-18 @-14 2-6 -#=7 - 








[crrA. 


XXIII. ] HARDER EQUATIONS. 149 


Solve the following equations : 
99 5a+31_ 22+9_ 42-6, 2e-13 
we t6 (#45 2-5 2-6 
12¢+1 Bo bia 2a 
Sa-1 1-922 1432 ~ 
es | ev 9 x—3 
a = hi eee 
24. x*7-9 x-3 xL+3 











28, 





[For additional examples see Hlementary Algebra. ] 


Literal Equations. 


181, In the equations we have discussed hitherto the co- 
efficients have been numerical quantities. When equations 
involve literal coefficients, these are supposed to be known, and 
will appear in the solution. 

Haample 1. Solve (x+a)(x+6)-—c(a+c) =(x-—c)(x+c)+ab. 

Multiplying out, we have 


x*+axn+ba+ab-—ac—c?=27-c?+ab; 





whence ax +bx = ac, 
(a+b)x=ac; 
ac 
Tete : 
a+b 
6b _a-b 





Hxample 2. Solve 


Simplifying the left side, we have 
a(x —b)—b(x-a)_a—b 


(e-a)(x—-b)  2aw-—c 








a 1 
(w-a)(x—b) xw-c 





Multiplying across, x?-—cwv =2?-ax-bxtab, 
ax+ba—cx = ab, 
(at+b-c)u=ab; 
ab 





150 ALGEBRA. 


Example 3. Solve the simultaneous equations : 





OL — DY A Cae, eset howto ea eee 
DEO =P Moe Eanes me ee 
To eliminate y, multiply (1) by g and (2) by 6; 
thus agqx — bay = cq, 
bpx + bay = br. 
By addition, (aq+bp)x =cq+br ; 
ey SE br 
aq +bp 


We might obtain y by substituting this value of x in ether of the 
equations (1) or (2); but y is more conveniently found by eliminat- 


ing x, as follows. 

Multiplying (1) by p and (2) by a, we have 
apx — bpy = cp, 
apx+aqy=ar. 

By subtraction, (aq+ bp)y = ar-cp; 
_ar-cp 

ay+ bp 





EXAMPLES XXIII, b, 


Solve the following equations : 





10, x+(2-a)(x—b)+a?+b0%? =b+27-a(b-1). 


ae ek ae 2 ys. ps 2 2 
vik Qu a_ 3x b_ 3a 8b 19. a-x b-2x a’ +b 














b a _ ab a—-b at ~ at oF 
ax—-b  bxe-c_a-cx x+a-b_x+b-c 
13. ¢ a a bai 14 xt+tb+e x+at+b 


ty a eee aN ey 
LO) nee q)*- pP(P a) + pa(Z 1) 0 





1, av+b?=a?- ba. 9, 2*-a? = (Qa-x), 
3. a@(a—x)+aba = b*(x—-b). 4, (b4+1)\(x%+a) =(b-1)(a-a) 
5, a(«+b)-b?=a?-b(a-2), 6. ca Ct=ad2+c. 
7, a(e—a)+b(a—b)+e(x-c) = 2(ab+be+ca). 
a? b? Qe) ee 
A es psy : = Fa eee, 
tbe aby Oxas i aa 


XXIII. ] HARDER EQUATIONS. Tol 


Solve the following simultaneous equations : 














6. x-y=a4+), 17, ce-dy=+a, Toe exer by, 
y 
ax + by = 0. w+ y= 2c. UC, 
tes ab _ EN hah 
19, pe ne 90. x y 0, yA a-y bv 
i Yn oe ¥-9, BELG 25) 
pon - ute a+b 
z_y_a,b etY _€-Y_¢, 
= Ss C= Ye PY So he 
a(a+2) = b(b-y) Dp ae 2 [on AGS 
94 2a—-b_2yta_su+y 95 axtby_1_ a? 
i a b a+2b * betay 2 bx+ay 


Irrational or Surd Equations. 


182. Derintrion. If the root of a quantity cannot be ex- 
actly obtained the indicated root is called a surd. 

Thus ./2, 9/5, /a3, Na?+0? are surds. 

A surd is sometimes called an irrational quantity; and 
quantities which are not surds are, for the sake of distinction, 
termed rational quantities, 

183. Sometimes equations are proposed in which the un- 
known quantity appears under the radical sign. For a fuller 
discussion of surd equations the student may consult the e- 
mentary Algebra, Chap. xxxi1. Here we shall only consider a 
few simple cases, which can generally be solved by the follow- 
ing method. Bring to one side of the equation a single radical 
term by itself: on squaring both sides this radical will disappear. 
By repeating this process any remaining radicals can in turn 
be removed. 


Example 1. Solve 2 Je —n/4e-11 = 1. 


Transposing, 2/2 —-1=/4x—11. 
Square both sides; then 4xv-4,/x+1=42—11, 
4,/x = 12, 


ee oy 


e='9: 


152 ALGEBRA. [CHAP. XXIII. 
Example 2. Solve 24 Vxe—5 = 13. 
Transposing, Na-5=11. 
Here we must cube both sides; thus 2-5=1331; 

whence x = 1336. 


Hxample 3. Solve 


6Je-ll_2Je+1 





3,/x 


Multiplying across, we have 


(6,/%-11)(,/7+6) = 




















Jxz+6- 


3r/2(2J/a+1)3 





that is, 62-11 ,/2+36/x - 66 = 6%4+38,/2, 
—11,/x+36,/x-3,/x = 66, 
Da Mace tits 
Vx =3; 
Me eee 
EXAMPLES XXIII, c, 
Solve the equations : 
L Neoe= 1. 0. NS — 2a = 7. 3. Ne=7 =e. 
4, oJe+1=3. ieee fhe = ae 
7 1-52 =3N 1-2. 8, 2N5x2-3-7/x=0. 
9, N4a?— 112-7 = 2-3. 10. 38V1-7x+422 = 5-62. 
ll. 14+ V2°—-322+72—-ll=z, 12) Nee wee) 
13, V4a+134+2,/a = 13. 14, 34+ V12z2—33 = 2A/32. 
15 Je-1_J/¢-3 16 Net _ J/e+5 
ie. ir ie ‘ 32-8 3\/2=7 
; ] 
17 Mepey ae 19, 2v%=7-9, 1 
Jxa-1 Wee Ja 14 4,/x-13 
3 3 
3 Sane 1 
Ee poecae, ly —-3=—_——. 
19. 1+474+2,/x AEs 90. Va+nNa-8 pe 
Q1. N4a+7-NaetrlsNe-3. 99, Nde-3-Nat+3= Ve-4. 


CHAPTER XXIV. 


HARDER PROBLEMS. 


184, Is previous chapters we have given collections of 
problems which lead to simple equations. We add here a few 
examples of somewhat greater difficulty. 


Example 1. If the numerator of a fraction is increased by 2 and 
the denominator by 1, it becomes equal to 8; and if the numerator 
and denominator are each diminished by 1, it becomes equal to 4: 
find the fraction. 


Let a be the numerator of the fraction, y the denominator ; then 


- . xv 
She fraction is =. 


From the first supposition, 





H+2_5 
yt ae) peor reerer eee eeeoeroee eeerreeoesses (Ly 
from the second, 
rd alt 
— = Fever sevescnes Peorereese ee evcoen 2, s 
ae (2) 


From the first equation, 8x#-d5y=-11, 
and from the second, 22—- y=1; 
whence x= 8, y=15. 


Thus the fraction is = 


Hxample 2. At what time between 4 and 5 o’clock will the 
minute-hand of a watch be 13 minutes in advance of the hour-hand ? 


Let x denote the required number of minutes after 4 o’clock ; 
then, as the minute-hand travels twelve times as fast as the hour- 


hand, the hour-hand will move over 5 minute-divisions in 2 minutes, 


a 


At 4 o’clock the minute-hand is 20 divisions behind the hour-hand, 
and finally the minute-hand is 13 divisions in advance; therefore the 
minute-hand moves over 20+13, or 33 divisions more than the hour- 
hand. 


154 ALGEBRA. [crar. 


a 
Hence soSe + 33, 
11 
—— i 33 ; 
12" 
Viera. 


Thus the time is 36 minutes past 4. 


If the question be asked as follows: ‘‘ At what times between 
4 and 5 o’clock will there be 13 minutes between the two hands?” 
we must also take into consideration the case when the minute-hand 
is 13 divisions behind the hour-band. In this case the minute-hand 
gains 20 - 13, or 7 divisions. 


Hence x = wat 7; 
which gives Sehr i 


, 
Therefore the times are i past 4, and 36’ past 4. 


Example 3. A grocer buys 15 Ibs. of figs and 28 Ibs. of currants 
for $2.60; by selling the figs at a loss of 10 per cent., and the cur- 
rants at a gain of 30 per cent., he clears 30 cents on his outlay ; 
how much per pound did he pay for each ? 


Let 2, y denote the number of cents in the price of a pound of 
figs and currants respectively ; then the outlay is 


154%-+28y cents. 
Therefore LODE 28 Y= 200 Fe eeitn se lawienaeatenes tee (3; 


The loss upon the figs is “6 x 15a cents, and the gain upon the 


currants is x 28y cents ; therefore the total gain is 


a) Te cents ; 
5 2 
pp eey 2h Tay 
EAD ; 
that is, BY = Daa 00g ag ete h en eee een (2). 


From (1) and (2) we find that «=8, and y=5; that is, the figs 
cost 8 cents a pound, and the currants cost 5 cents a pound. 


Example 4. Two persons A and B start simultaneously from 
two places, ¢ miles apart, and walk in the same direction. A travels 
at the rate of p miles an hour, and B at the rate of g miles ; how far 
will A have walked before he overtakes B ? 


XXIV, ] HARDER PROBLEMS. 155 


Suppose A has walked x miles, then B has walked a —c miles. 


A walking at the rate of p miles an hour will travel 2 miles in 


x—-—C 


* hours; and B will travel a—c miles in hours; these two 
) 





times being equal, we have 





a _m-¢ 
fo) ee 
Gx = px—pe 5 
whence peed San 
Ped 
Therefore A has travelled ae miles. 
9) coat 


Example 5. A train travelled a certain distance at a uniform 
rate. Had the speed been 6 miles an hour more, the journey would 
have occupied 4 hours less ; and had the speed been 6 miles an hour 
less, the journey would have occupied 6 hours more. Find the 
distance. 


Let the speed of the train be x miles per hour, and let the time 
occupied be y hours ; then the distance traversed will be represented 
by zy miles. 

On the first supposition the speed per hour is 7 +6 miles, and the 
time taken is y—4 hours. In this case the distance traversed will 
be represented by (x+6)(y- 4) miles. 

On the second supposition the distance traversed will be repre- 
sented by (x—6)(y+6) miles. 

All these expressions for the distance must be equal ; 


xy = («+6)(y —4) = (w@-6)(y+6). 
From these equations we have 
xy = xy + by — 4x - 24, 


or Oran aerlee aes itn. gnaw an tanhihrare eee: Cl ys 
and xy = xy — by + 6x — 36, 
or Goes Ope VOM ate ae seein en'n.s 8016 Riss anions (2). 


From (1) and (2) we obtain x = 30, y = 24. 
Hence the distance is 720 miles. 


EXAMPLES XXIV. 


1, If the numerator of a fraction is increased by 5 it reduces to 4, 
and if the denominator is increased by 9 it reduces to 4: find the 
fraction. 


156 ALGEBRA. [cHAP., 


9, Find a fraction such that it reduces to 2 if 7 be subtracted from 
its denominator, and reduces to 2 on subtracting 3 from its numerator. 


3. If unity is taken from the denominator of a fraction it reduces 
to 4; if 3 is added to the numerator it reduces to +: required the 
fraction. 


4, Find a fraction which becomes # on adding 5 to the numerator 
and subtracting 1 from the denominator, and reduces to 4 on sub- 
tracting 4 from the numerator and adding 7 to the denominator. 


5, If 9 is added to the numerator a certain fraction will be 
increased by 3; if 6 is taken from the denominator the fraction 
reduces to 2: required the fraction. 


6. At what time between 9 and 10 o’clock are the hands of a 
watch together ? 


7, When are the hands of a clock 8 minutes apart between the 
hours of 5 and 6? 


8, At what time between 10 and 11 o’clock is the hour-hand six 
minutes ahead of the minute-hand ? 


9, At what time between 1 and 2 o’clock are the hands of a 
watch in the same straight line ? 


10, When are the hands of a clock at right angles between the 
hours of 5 and 6? 


11, At what times between 12 and 1 o’clock are the hands of a 
watch at right angles ? 


12, A person buys 20 yards of cloth and 25 yards of canvas for 
$35. By selling the cloth at a gain of 15 per cent. and the canvas 
at a gain of 20 per cent. he clears $5.75 ; find the price of each per 
yard. 

13, <A dealer spends $1445 in buying horses at $75 each and 
cows at $20 each ; through disease he loses 20 per cent. of the horses 
and 25 per cent. of the cows. By selling the animals at the price 
he gave for them he receives $1140; find how many of each kind 
he bought. 


14, The population of a certain district is 33000, of whom 835 
can neither read nor write. These consist of 2 per cent. of all the 
males and 3 per cent. of all the females: find the number of males 
and females. 


15, Two persons C and D start simultaneously from two places ~ 
a miles apart, and walk to meet each other; if C walks p miles per 
hour, and D one mile per hour faster than C, how far will D have 
walked when they meet? 


16, A can walk a miles per hour faster than B; supposing that 


he gives B a start of c miles, and that B walks n miles per hour, 
how far will 4 have walked when he overtakes B? 


XXIV. | HARDER PROBLEMS. TOW 


17, A, B, Cstart from the same place at the rates of a, a+b, 
a+2b miles an hour respectively. B starts n hours after A, how 
long after B must C start in order that they may overtake A at the 
same instant, and how far will they then have walked ? 


18. Find the distance between two towns when by increasing 
the speed 7 miles per hour, a train can perform the journey in 1 
hour less, and by reducing the speed 5 miles per hour can perform 
the journey in 1 hour more. 


19, A person buys a certain quantity of land. If he had bought 
7 acres more each acre would have cost $4 less, and if each acre 
had cost $18 more he would have obtained 15 acres less: how much 
did he pay for the land ? 


90, Acan walk half a mile per hour faster than B, and three- 
quarters of a mile per hour faster than C. To walk a certain dis- 
tance C takes three-quarters of an hour more than B, and two 
hours more than A: find their rates of walking per hour. 


91, A man pays $90 for coal; if each ton had cost 50 cents 
more he would have received 2 tons less, but if each ton had cost 
75 cents less he would have received 4 tons more ; how many tons 
did he buy ? 


99, Aand B are playing for money; in the first game A loses 
one-half of his money, but in the second he wins one quarter of 
what Bthen has. When they cease playing, A has won $10, and 
B has still $25 more than A; with what amounts did they begin ? 


93, The area of three fields is 516 acres, and the area of the 
largest and smallest fields exceeds by 380 acres twice the area of 
the middle field. If the smallest field had been twice as large, and 
the other two fields half their actual size, the total area would 
have been 42 acres less than it is; find area of each of the fields. 


94, A, B, Ceach spend the same amount in buying different 
qualities of cloth. B pays three-eighths of a dollar per yard less 
than A and obtains three-fourths of a yard more; C pays five- 
eighths of a dollar per yard more than A and obtains one yard less ; 
how much does each spend ? 


95, B pays $28 more rent for a field than A; he has three- 
fourths of an acre more and pays $1.75 per acre more. C pays 
$72.50 more than A; he has six and one-fourth acres more, but pays 
25 cents per acre less ; find the size of the fields. 


158 ALGEBRA. [ciAP. 


MISCELLANEOUS EXAMPLES IV. 


1. When a=-—3, 6=5, ¢=—1, d=0, find the value of 
26c¢ ab — Cd +5be —4ac + a2. 





9. Solve the equations : 


Gee's! i Fanless 
(1) ey ae a aatg pl 88 32933 


(2) l=y+2=2(2+2) = 3(v+y). 


3, Simplify 

ty Gee ee 
Cele Ole on x—-a 
b2 — 3b 2 i? +b —30 aL ob: = Ob 
b?-2b+4 b?+3b-18 Pou is 





(2) 





4, Find the square root of 


1bO SO 
~ 36x + 602? — 3 yA 
9 x+602 3 x 5 





5, Ina base-ball match the errors in the first four innings 
are one-fourth of the runs, and in the last five innings the errors 
are one-third of the runs. The score is 16, and the errors num- 
ber 5; find the score in the first four innings. 


1 


Pe 8 
G) Vind the value ote (ee 


~J 


. Find the value of 


I 


1 ST Pah ss Liat 
(a+2)-3(1- 2p) - (20-3043) +3p-4(ta~2), 


Gol to 


8, Resolve into factors 
(1) 38a?-20a-7; (2) atb?- dita? 


AP fa ee 


9, Reduce tolowest terms ~~. rs Est, 
4x3 + 522 -—Tx-2 


XXIV. | MISCELLANEOUS EXAMPLES IV. 159 


10, Solve the equations: 


Pe 8 AR po Sr 
(1) w-3 3 o rt 


(2) x+y-2=0, w-ytz=4, 5xe+y+2=20; 


(3) axt+b, dxte os 


¢ a 


ee, 3 x +2 4 
mate CES caer, 
UG MEMOUUA 6 bc ger erga cages my Te mam OT] 











12. A purse of sovereigns is divided amongst three persons, the 
first receiving half of them and one more, the second half of the 
remainder and one more, and the third six. Find the number of 
sovereigns the purse contained. 


13, If h=-1, &=2, 1=0, m=1, n=-3, find the value of 
h3(m —l) - N3hn+hk 
m(l—h) —2hm? + Vthk 
14, Find the L.C.M. of 
15(p3+q°), 5(p?- pqt+q’), Mp? +pq+@), 6(p?- 9’). 
15, Find the square root of 


nae 9g 
LS 
(1) 9 ye 


(2) 1~6a+5a2-+ 1203+ 4a4, 








: , 20a?+27%+9 20x74 277+9 
_  Simplif 
Soe Opty cide 12a? +17x+6 


17, Solve the equations: 





ao o)e bie aye 
Dee ah hie Fees 
(2) 9 ye 3 4 8 


Z—4 Hoan geo 3. 


18, A sum of money is to be divided among a number of per- 
sons; if $8 is given to each there will be $3 short, and if $7.50 is 
given to each there will be $2 over: find the number of persons. 


160 ALGEBRA. [cuar. 


19, Resolve into factors : 
(1) 2a?-3ab+(a-—6bd)z ; (2) 4a?-4ay— 15y?. 


20, In the expression 2° —2a?+3x-4, substitute a—-2 for 2, and 
arrange the result according to the descending powers of a. 


21, Simplify 


x x 
NR a 3 OS rere 
ie peas 
ae ie Se 
a 


O2>) Hind: the WH, ©. ot 
323 —1le?+a%+4+15 and 5a4— 7x3 — 20a? - lla -3. 
23, Express in the simplest form 


See 


he at 3 
C1y ee eee e eee ee 
LEY Ee e-l1 «+1 x + 


y x 


24, <A person possesses $5000 stock, some at 3 per cent., four 
times as much at 34 per cent., and the rest at 4 per cent. : find the 
amount of each kind of stock when his income is $176. 


25, Simplify the expression 
~ 3[(a-+b) — {(2a — 3b) - (5a +7b —16c) - (- 13a+ 2b - 8c - 5d)}}, 
and find its value when a= 1, b= 2)¢=.o.8@.=14. 
26, Solve the following equations : 
(1) ligsea10, die—10ly—110; 
(2) wty-2=3, w+2-y=5, y+2-x=7 


27, Express the following fractions in their simplest form : 








32x? — 24412 1 
pes a 2 —_— 
) 12a> — u4+427’ (2) Bote! 

1, 2+3 
ates 
pa: 


98, What value of a will make the product of 3-8a and 8a+4 
equal to the product of G6a+11 and 3-4a? 


XV] MISCELLANEOUS EXAMPLES IV. 161 


99, Find the L.C.M. of 2? -—2?-3x-—9 and 2° - 2z?-5x2—- 12. 


30, A certain number of two digits is equal to seven times the 
sum of its digits: if the digit in the units’ place be decreased by 
two and that in the tens’ place by one, and if the number thus 
formed be divided by the sum of its digits, the quotient is 10. Find 
the number, 


81, Find the value of 


6a? — Bay — Gy? . 3x? — xy — 4 2, 9a" — Gay — Sy" 
2x? +ay—y? Qa? — Bayt dy?” La —Bayt+y? 








82. Resolve each of the following expressions into four factors : 
(1) 4a4—17a?b? + 404; (2) a8 —256y8. 


383, Find the expression of highest dimensions which will divide 
24a4b — 2a%b? — 9ab* and 18a° + a4b? —- 6a°b*? without remainder. 


34, Find the square root of 

(I) w(a+1)(e+2)(4+3)+1; 

(2) (2a?+13a+15)(a?+ 4a —5)(2a2+ a - 3). 
35, Simplify 


36. A quantity of land, partly pasture and partly arable, is sold 
at the rate of $60 per acre for the pasture and $40 per acre for the 
arable, and the whole sum obtained is $10000. If the average price 
per acre were $50, the sum obtained would be 10 per cent. higher: 
find how much of the land is pasture, and how much arable. 


CHAPTER XXV. 
QUADRATIC EQUATIONS. 


185, Derinirion. An equation which contains the square 
of the unknown quantity, but no higher power, is called a quad- 
ratic equation, or an equation of the second degree. 

If the equation contains both the square and the first power 
of the unknown it is called an affected quadratic; if it contains 
only the square of the unknown it is said to be a pure quadratic. 

Thus 2x?7—5xz=3 is an affected quadratic, 
and 5x2=20 is a pure quadratic. 


Pure Quadratic Equations. 


186, A pure quadratic may be considered as a simple equa- 
tion in which the square of the unknown quantity is to be found. 


Example. Solve ——— Hiatal — ei. 
v?—-27 2-11 
Multiplying across, 9x?—99=25x?— 675 ; 
Hak Veni Jaa yh 
ios ob 5 
and taking the square root of these equals, we have 
L= +6. 
[In regard to the double sign see Art. 119. ] 


187. In extracting the square root of the two sides of the 
equation z?=36, it might seem that we ought to prefix the 
double sign to the quantities on both sides, and write +z2=+6. 
But an examination of the various cases shows this to be un- 
necessary. For +x=+6 gives the four cases: 

t+z4=+4+6, +x=—-6, —x7=+6, —x=—6, 
and these are all included in the two already given, namely 
x=+6,2=—6. Hence, when we extract the square root of the 


two sides of an equation, it is sufficient to put the double sign 
before the square root of one side. 


CHAP, XXV. | QUADRATIC EQUATIONS. 163 


Affected Quadratic Equations. 


188. The equation z?=36 is an instance of the simplest form 
of quadratic equations. The equation (#—3)?=25 may be 
solved in a similar way; for taking the square root of both 
sides, we have two simple equations, 


Bao 5, 

Taking the upper sign, #2—3=+5, whence 7=8; 

taking the lower sign, x—3=—5, whence r= — 2. 

.. the solution is = 6,06 2. 

Now the given equation (7—3)?=25 

may be written x* — 6x +(3)?=25, 

or x? —62=16. 

Hence, by retracing our steps, we learn that the equation 
w= 6216 


can be solved by first adding (3)? or 9 to each side, and then 
extracting the square root; and the reason why we add 9 to 
each side is that this quantity added to the left side makes it a 
perfect square. 
Now whatever the quantity a may be, 

a+ 2ar+a°=(¢@+a)’, 
and uv —2ax+a?=(«4-ay; 
so that if a trinomial is a perfect square, and zits highest power, 
x*, has unity for its coefficient, we must always have the term 
without w equal to the square of half the coefficient of w. If, 


therefore, the terms in xz? and z are given, the square may be 
completed by adding the square of half the coefficient of vr. 


Example. Solve 2?+14x = 32. 
The square of half 14 is (7)?. 


that is, (fF 7 j= 81 ; 


189, When an expression is a perfect square, the square terms 
are always positive. Hence, before completing the square the 
coefficient of x” should be made cqual to +1, 


164 ALGEBRA. [orar. 


Hxamplel. Solve Fx =2?-8. 


Transpose so as to have the terms involving x on one side, and the 
square term positive. 


Thus ele = 8. 
Completing the square, 2? -—7a+ & = oe : 
that is, Ga ; 
2 + 
4 bis iin 2 
pee), 
odio > 
Ze 


$  _ 8x45 
38a+1 3x2+1- 





Example 2. Solve 4- 


Clearing of fractions, 12%+4-8 = 327+5; 
bringing the terms involving 2 to one side, we obtain 
3x7 — 12% =-9. 
Divide throughout by 3; then 


x?-4e=-3; 
x? -444+(2)?=4-3; 
that is, (w= Que 
e-2=+1; 
Gis. OL ad 


EXAMPLES XXV. a. 


Solve the equations : 


Ll.) W(=7) = 627. 2. (24 8)(e—-8)=17. Be (i eal aye 





eee oe eerie ee an” 
eae Drea: 8, 2?+6a = 40. = Baa gale 4a 

10) 927 £25. Lie? 1G =a: Loe Leia 

18, 27+47=32. 14, 974+36=27. 15, 2«7+152-34=0. 


XXV.] QUADRATIC EQUATIONS. 165 


Solve the equations : 











] a9 J eb) 2 2 9 tA _ “+37 
16, ee ee) Ne ae +3). Lis ane ei 
+3, %-2_ 5 Deaar Le near ge 
18. e2+2 “2-3 (#+2)(x—3) 19. Qa +2 eri 
2 Se eee 
90, 5 aay x+2, 


190. We have shown that the square may readily be com- 
pleted when the coefficient of #2 is unity. All cases may be 
reduced to this by dividing the equation throughout by the 
coefficient of 2°. 


Example 1. Solve 32-3827=10z. 





Transposing, 3x? + 10% = 32. 
Divide throughout by 3, so as to make the coefficient of x nity. 
oy ANU Sy 
TI Co . 
lus 3 op 3 
5\2 OF 
Completing the square, 27 + Put 3 a4 ; 
ae oy 4h. 
that is, (« oe rg 
5) 11 
C+—= *-— 3 
oS 
e 
a3 =2, or —5% 


Example 2. Solve 527+1llx=i2. 





Dividing by 5, eo AZ aae 
Completing thesquare, 4 1 Gar = + “i - 
that is, (2+ i = ae ; 
ete 


166 ALGEBRA. [cIAP. 


191. We see then that the following steps are required for 
solving an affected quadratic equation : 


(1) If necessary, simplify the equation so that the terms in 
x“ and x are on one side of the equation, and the term without x 
on the other. 

(2) Make the coefficient of x* unity and positive by dividing 
throughout by the coefficient of x”. 

(3) Add to each side of the equation the square of half the 
coefficient of x. 

(4) Take the square root of each side. 

(5) Solve the resulting simple equations. 


192. When the coefficients are hteral the same method may 
be used. 


Hxample. Solve 7(#+2a)*+ 8a? = 5a(7x + 23a). 
Simplifying, 1x7 + 28ax + 28a? + 8a? = 35ax + 11da? ; 





that is, 7x? — Tax =84a?, 
or Tae Ce OO. 
Completing the square, 2*-ax+ OF = 12a + ; 
2 2 
that is, (« oe B = ae 3 
C— x = +14 R 
2 2 
x=4a, or —3a 


193, In all the instances considered hitherto the quadratic 
equations have had two roots. Sometimes, however, there is 
only one solution. Thus if #?—-227+1=0, then (7—1)’=0, 
whence w=1 is the only solution. Nevertheless, in this and 
similar cases we find it convenient to say that the quadratic has 
two equal roots. 


EXAMPLES XXV., b, 
Solve the equations : 
1, 3a2+2a = 21. 9, 5a2=S8a42]. 8 6a2-x-1=0. 
43-1 le =e"; He 2a = Dares. 6, 10+232+4 122? = 0. 
7, 152?-6x = 9. §, 4a0°-17a = 15. Q, 8a?-197-15=0. 


XXV. J QUADRATIC EQUATIONS, 167 


Solve the equations : 
HH) 1077 oe = 1. Pee Ta = 12: 12, 2027—2-—1=0. 
toe pean loa. 14° 29*—S8at= lbax. 15, 3a*=k(2k— 5a). 
nGseloee 20e ans 17, 9a" — 14h = box, 18. 2e72* = ar). 
19, (w7-3)(~—-2) = 2(a7-4). 90, 5(%+1)(8%+4+5) = 3(82?+ lla +10). 
D1, 322+134+ (%-1)(2%4+1) = 2x(2x 4+ 3). 





Tx-3_ 3x 2 2-] 3a-1_2a-9 
22, a O° 23, 375577 24, eto ei 4 
62-5 2 x-4 1 11 
== Oo. See Shed oot Se ec 
25. z+5 3 26, je+1 2 2(3+ 22) 


27, 3(2a+ 3)? + 2(2x% +3)(2- a) = (w- 2)", 
98, (38a —7)? — (2a - 3)? = (w—4)(8a+1). 
[For additional examples see Elementary Algebra. | 


194, Solution by Formula, From the preceding examples 
it appears that after suitable reduction and_ transposition 
every quadratic equation can be written in the form 


an +br+c=0, 


where a, 6, c may have any numerical values whatever. If 
therefore we can solve this quadratic we can solve any. 


Transposing, ax’ +be=—c¢; 
anon b c 
dividing by a, Cg, 
a a 


Complete the square by adding to each side cal ; thus 
20 


faa nee bee, 
a 








2a Aa? a’ 
: b\2 6b%—4ae 
that is (« ey = s 
; oe 4a? 
extracting the square root, 
+ Ge 





2a 2a 
bie Ot (Ot 4ac) 
Qa ; 


168 ALGEBRA. [CHAP. 


—b+ ,/(b’ — 4ac) 
2a 4 


it must be remembered that the expression ,/()?—4ac) is the 
square root of the compound quantity b? —4ac, taken as a whole. 
We cannot sumphfy the solution unless we know the numerical 
values of a, b,c. It may sometimes happen that these values 
do not make b?—4ac a perfect square. In such a case the exact 
numerical solution of the equation cannot be determined. 


195. Inthe result w= 





Hxample. Solve 52?-13x-11=0. 
Here a=5, b=-13, c=~—11; therefore by the formula we have 


(—13) + \(S19P—4 7) B11) 











oa 
2.5 
_13 + V169 +22) 
10 
_ 13+ /389 





10 
Since 389 has not an exact square root this result cannot be 


simplified ; thus the two roots are 


13+ ,/389 13— ,/389 
102s 10 





196. Solution by Factors. There is still one method of 
obtaining the solution of a quadratic which will sometimes be 
found shorter than either of the methods already given. 


Consider the equation 2+ la = 
Clearing of fractions, $3274 72-60 assent -ceecesnessennaee CLs 


by resolving the left-hand side into factors we have 
(3x2 — 2)(a+3)=0. 
Now if either of the factors 32 —2, «+3 be zero, their product is 
zero. Hence the quadratic equation is satisfied by either of the 
suppositions 
a7 —-2=0, or r+3=0. 
> Ef) 

It appears from this that when a quadratic equation has been 
simplified and brought to the form of equation (1), its solution 
can always be readily obtained if the expression on the left-hand 


Thus the roots are 


XXv.] QUADRATIC EQUATIONS. 169 


side can be resolved into factors. Hach of these factors equated 
to zero gives a simple equation, and a corresponding root of the 
quadratic. 


Example 1. Solve 2x°-ax+2bx = ab. 


Transposing, so as to have all the terms on one side of the equation, 
we have 
2x7 —ax+2bxe—-ab=0, 


Now 22:7 — ax + 2ba -—ab = x(2% -a)+b(2x%-a) 
= (2%-a)(x+b). 
Therefore (Qe-a)\(v+l)=0; 
whence 22-—-a=0, or 7+b=0, 


ane, or —b. 
Y 


Example 2. Solve 2(x?-6) = 3(a”—-4). 


We have 2e2 \2 = oe = 12 > 
that is, CA laa Woe ee RAPE APT tee NORA (1). 
Transposing, 20% Bu =), 
x(2%—-3)=0 
v= O,or 27 = 3 =0 


Thus the roots are 0, 2 


Note. In equation (1) above we might have divided both sides by x 


a] 
3 . > oO . F, 
and obtained the simple equation 2a = 3, whence x = 5, which is one 


of the solutions of the given equation. But the student must be 
particularly careful to notice that whenever an x is removed by 
division from every term of an equation it must not be neglected, 
since the equation is satisfied by x=0, which is therefore one of 
the roots. 


197, Formation of Equations with given roots, It is 
now easy to form an equation whose roots are known. 


Example 1. Form the equation whose roots are 4 and —3, 
Here : v= Ay or asia) * 
G=4'3Q, Of T+ a= 0 3 
both of these statements are included in 
(x —4)(x+3) =0, 
or a*-x2-12=0, 
which is the required equation, 


170 


ALGEBRA. 


[CHAP. 


Example 2. Form the equation whose roots are a and - : : 


Here 


the equation is 


that is, 
or 320? 


(x —a)(8a+ 6) = 0, 
—8axr+ba-—ab=0. 


EXAMPLES XXV. c. 


Solve by formula the equations : 











a oi Uk —24-1=0. 3. x7 ore: 
Ae Oar =e eel 5, Qa?-9x = 4. 6. 327+7x2=6. 
7 4a2—14 = 82. 8. 6a2-3-—Ta=0. 9, 1222410 = 23m. 
Solve by resolution into factors : 
10.30 —92 =90; Teale] 1625 19) ar 8a = 1 
13, 2_ 37 = 2, 14) 3224+52-+42=0"" 15, 427° ite 
16.6 Ser Sle 2 0) 17 a2 — 0.0; 182° 2? = Tax =8e% 
19, 122? — 2362 + 100? = 0. 90, 3ax?+2ba = Tax. 
91, 2422+ 22cx = 21c?. 09. a — 22 44h = 2he, 
Solve the equations : 
93, 2a(a+9) =(e%4+1)(5--z). 
94, (2x-1)?-11 = 5x+(x-3)2. 
95, G(x -—2)?+138(1 - x)(a - 2) + 6a? = 6(2x -1). 
= 4 3 2 2. 
aan 9, EASA = 
26, x-6 x-5 27. Sx-1l x+1 2 
10 oe +3 4(x~6)_ 
a a ese se ea Oo ae ie 
] Gon Wines 
eee sat as 
a Sseeeig S13 Mead} 
Tee OG hs TE) e415 
Sakon teal apres ee ao. 
32, io ales wre 


Xxv.] QUADRATIC EQUATIONS. 171 


Solve the equations : 








eee et" oot 

34, Gry Vel Oa” 35. Bar’ — 4 aap 

Ce eee — 3 Pel D) u ei? 

Hi og oT el eA 
3, SEM a G2 99, (p— abe + 2 = Wn +9. 

b P 5b c\? 

2 aN a 

40, Spans 41. i 1) (: i) ® 


[For additional examples see Hiementary Algebra. | 


198. Simultaneous Quadratic Equations. If from either 
of two equations which involve w and y the value of one of the 
unknowns can be expressed in terms of the other, then by sub- 
stitution in the second equation we obtain a quadratic which 
may be solved by any one of the methods explained in this 
chapter. 


Example. Solve the simultaneous equations 
5xt+Ty=1, 447+3axy —2y?=10. 





From the first equation, x = a , and therefore by substitution 


in the second equation, we have 
a(S ures ye 272=10; 





25 
whence 4 — 56y + 196y? + 15y — 105y? — 50y? = 250 ; 
that is, 4ly*—4ly —246=0 ; 
| y’-y-6=0; 
(y—8)(y+2)=05 
y=3, or —2. 


From the first equation, we see that if y=3, then x= —4, and if 
y=—2, then x=. 


Homogeneous Equations of the Same Degree. 


199, The most convenient method of solution is to substitute 
y=me in each of the given equations. By division we eliminate 
x and obtain a quadratic to determine the values of m. 


ve ALGEBRA. (CHAP. XXV. 


Example. Solve the simultaneous equations 
527 + 3y?= 32, “x? -a2y +2y?=16. 
Put y= mez and substitute in each equation. Thus 
roms MaRS (Ed beciea ors Pare, a Pe SM eT A, oe | (qj 
and col 22) = LG eae eee ee 2). 
By division, Maweisil ayia 2 
l1-m+2m* 16 
that is, m* -2m-3=0; 
(m—3)(m+1)=0; 
= 3,.0F «2.1, 


(1) Take m=3 and Syustitnte in either (1) or (2). 


From (1), 32x? = 32: whence x= + 1. 
WA = a a es 





(2) Take m= —1 and substitute in (1). Thus 
827=32; whence v= + 2. 
Y=MNE= Lf. 2. 


EXAMPLES XXV. d. 


Solve the simultaneous equations : 


1, w+3y=9, 9, 38x-—4y=2, oe 2r+y=5, 
vy bs iy =e De — wy = 2. 
ee Te aaa 5 ot er vans, 6, 28-215 
x? + 4y? = 29. 3xy — y= 9. mete Lb 
co on ae ies) ai pot 
hs aa 1; 8, Batry fs ’ 9, x oe 
xy = 24 10zy=1 xy —yr=4 
eel eat Pees be 
ieee . — A Lie Dates itm 
10.5 5 Le ong ae ; 
3.Y ee) seein 
Prv=] peas a | — = 8, 
22 ee ey oe 
10; 327-79? 255... 14, léry—-Se*= 77, 15. a2 ey a 
2c? ey OU. Tay + 3y? = 110. Say + 3y7 = 195. 
16, 27+ 2ay + 2y? = 17, 17, 2la?+3ay - y? =37)1, 


3a? — Oxy — y= 119. 5a? + day + Sy? = 265. 


CHAPTER XXXVI. 


PROBLEMS LEADING TO QUADRATIC EQUATIONS. 


200. Wr shall now discuss some problems which give rise 
to quadratic equations. 


Hxample 1. A train travels 300 miles at a uniform rate; if the 
speed had been 5 miles an hour more, the journey would have taken 
two hours less: find the rate of the train. 


Suppose the train travels at the rate of x miles per hour, then the 


: See ay 
time occupied is —— hours. 
i 


ry 





On the other supposition the time is a5 hours ; 
5 
300 _300_o. 
Se Ce ee ’ 
ce oe ie 
whence e+ 5a —750 = 0, 
or (7+ 30)(% — 25) = 0, 
* oa = 25,06) — 30. 


Hence the train travels 25 miles per hour, the negative value 
being inadmissible, 

{For an explanation of the meaning of the negative value see 
Elementary Algebra. | 


EHxrample 2. Aman buys a number of articles for $2.40, and sells 
for $2.52 all but two at 2 cents apiece more than they cost; how 
many did he buy ? 


Let « be the number of articles bought; then the cost price of 





each is ae cents, and the sale price is 5 cents. 
HY Spas 
252 240_.. 
fo ee f 
that is, Oe me Eas 





174 ALGEBRA. [cHAP. 


After simplification, 6x +240 = a? - 2a, 
or x? —8x2-240=0; 
that is, (7 —20)(4+12)=0; 


Lia UP Orie, 


Thus the number required is 20. 


Example 3. A cistern can be filled by two pipes in 334 minutes ; 
if the larger pipe takes 15 minutes less than the smaller to fill the 
cistern, find in what time it will be filled by each pipe singly. 


Suppose that the two pipes running singly would fill the cistern 





in x and «-15 minutes; then they will fill ba i yet 5 of the cistern 
2 


a 


respectively in one minute, and therefore when running together they 





will fill (+ i of the cistern in one minute. 
But they fill a , or a of the cistern in one minute. 
1 1 3 
H = eee niet 
a fen anOG 


100(2% - 15) = 38x(x—- 15), 
3a? — 245x + 1500 =0, 
(a —75)(3a —20)=0; 
x ='75, or 62. 
Thus the smaller pipe takes 75 minutes, the larger 60 minutes, 


The other solution 62 is inadmissible. 


901. Sometimes it will be found convenient to use more than 
one unknown. 


Example. Nine times the side of one square exceeds the peri- 
meter of a second square by one foot, and six times the area of the 
second square exceeds twenty-nine times the area of the first by one 
square foot; find the length of a side of each square. 


Let x feet and y feet represent the sides of the two squares ; then 
the perimeter of the second square is 4y feet ; thus 


9x. 4y-= 1, 
The areas of the two squares are x” and y” square feet; thus 
6y? - 2927 = 1, 


XXVI.] PROBLEMS LEADING TO QUADRATIC EQUATIONS. 175 


From the first equation, Ye a _ 





By substitution in the second equation, 





Se Woe oon 1 : 
8 
that is, llz?-—547%-5=0, 
or (2—d5)(llz+1)=0; 
whence x =5, the negative value being inadmissible. 
Also, y = a L STi 


Thus the lengths are 5 ft. and 11 ft. 


EXAMPLES XXVI, 


], Find a number which is less than its square by 72. 


9. Divide 16 into two parts such that the sum of their squares 
is 130, 


3, Find two numbers differing by 5 such that the sum of their 
squares is equal to 233. 


4, Find a number which when increased by 13 is 68 times the 
reciprocal of the number. 


5, Find two numbers differing by 7 such that their product 
is 330. 

6, The breadth of a rectangle is five yards shorter than the 
length, and the area is 374 square yards: find the sides. 


7, One side of a rectangle is 7 yards longer than the other, and 
its diagonal is 13 yards ; find the area. 


8, Find two consecutive numbers the difference of whose reci- 
procals is sz. 

9, Find two consecutive even numbers the difference of whose 
reciprocals is <3. 


10. The difference of the reciprocals of two consecutive odd 
numbers is 725: find them. 


11, A farmer bought a certain number of sheep for $315; 
through disease he lost 10, but by selling the remainder at 75 cents 
each more than he gave for them, he gained $75: how many did 
he buy ? 

12, By walking three-quarters of a mile more than his ordinary 
pace per hour, a man finds that he takes 1} hours less than usual to 
walk 293 miles; what is the ordinary rate ? 


176 ALGEBRA. (CHAP. XXVI. 


13, <A cistern can be filled by the larger of two pipes in 5 min- 
utes less than by the smaller. When the taps are both running the 
cistern is filled in 6 minutes: find the time in which the cistern 
could be filled by each of the pipes. 

14, A man buys a dozen eggs, and calculates that if they had 
been a cent per dozen cheaper he could have bought two more for 
twelve cents: what is the price per dozen ? 

15, The large wheel of a carriage is one foot more in circum- 
ference than the small wheel, and makes 48 revolutions less per 
mile: find the circumference of each wheel. 


16. A boy was sent out to buy 12 cents’ worth of apples. He 
ate two, and his master had in consequence to pay at the rate of a 
cent per dozen more than the market price. How many apples did 
the boy buy ? | 

17, A lawn 45 feet long and 40 broad has a path of uniform 
width round it; if the area of the path is 50 square yards, find its 
width. 

18. By selling one more apple for a cent than she formerly did, 
a woman finds that she gets a cent less per dozen: how much does 
she now get per dozen ? 

19, Four times the side of one square is less than the perimeter 
of a second square by 12 feet, and eleven times the area of the first 
is less than five times the area of the second by 9 square feet : find 
the length of a side of each square. 

20, Find a number of two digits such that if it be divided by the 
product of its digits the quotient is 7, and if 27 be subtracted from 
the number the order of the digits is reversed. [Art. 111.] 

91, A person buys some 5% per cent. stock; if the price had 
been $5 less he would have received cone per cent. more interest on 
his money : at what price did he buy tLe stock ? 

92, The area of each of two rectangles is 1008 square feet ; the 
length of one is 8 feet more than that of the other, and the difference 
of their breadths is 3 feet: find their sidcs, 

23. There are three numbers of which the second is greater than 
the first by 6 and less than the third by 9. If the product of all 
three is 280 times the greatest, find the numbers. 


24, Find four consecutive integers such that the product of the 
two greatest is represented by a number which has the two least for 
its digits. 

95, Two trains A and B start simultaneously from two stations 
Pand Q which are 260 miles apart. A reaches Y in 32 hours, and 
B reaches P in 42 hours after they meet: find the rate of cach train. 


MISCELLANEOUS EXAMPLES V. 


[The following examples are arranged progressivel ye 1-24 may 
be taken after CHAP. XIII. ; 25-36 after Cuap, Xvit. ; 37-48 after 
CHap. xx. ; 49-60 after Cuap. xxit.; 61-72 after Crap. XXIV. 
The remaining examples are quite general and cover the contents 


of the whole book.) 


1, lf x=3, y=-2, 2=0, find the value of 
Bar? + Byz + 4y? 
cays 
2. Divide 3p>+ 16p*- 33p°+14p? by p?+7p. 
3, Find the sum of a-2(b-3c), 3{a-2(b+c)}, 2{b -2(a —2b)}. 
4, Simplify by removing brackets 7[8a-4{a-—b+3(a+b)}}. 
5, Solve the equations : 
x+4,«a-4 


(1) agentes =4; 


6. A is three times as old as B; two years ago he was five times 
as old as B was four years ago : what is A’s age ? “i 


7. Find the product of 2a -3b-(a-—2b-—c) and b-2c-—(a-c). 
Gide] b= 0, ¢=-11) do =2, ei 2) find the value of 
+4+0+E+4+0R—-8+a24+l2+C°+d?-e. 
9, Remove brackets from the expressions : 
(1) a -[5b-{a—(3c -3b)+2c- (a - eel 
(2) Qa - 3{b-4(c - d)}]-[a—4fb - 6(c--d)}]. 


10, If the price of 5 acres of land is $a, what is the price of x 
acres ? and how many acres can be bought for $b ? 

11, Divide at—4 by a?—2a42. 

12, There are 150 coins in a bag which are either half-dollars 
or quarters. If the value of the coins is $58.50, find the number 
of each kind. 


H.A. M 


178 ALGEBRA. 


13. Add together a—{b+c-(a+b)!+c, 2(3a+2b) -4(b+2a)-c, 
and 3(2b — a) -2(3b-a)+e. 


14, Find what value of x will make the product of +3 and 
2%+3 exceed the product of x+1 and 2x+1 by 14. 


15, Divide b?+8-125c?+30bc by b-5c+2. 
16, *Simplify 

(1) 13ab*c? ‘ 87c°d? , 26bed , 
29c4d ~ 2a2bt * 4ab3 ’ 
202 ee a®bx , atb%a4 ) 
“b2 \2Qab 2c? Ga2b — Qakb4x3/° 


17, How old will a man be in m years who n years ago was 
p times as old as his son then aged 2 years ? 








(2) 





18, I bought a certain number of pears at three for a cent, 
and two-thirds of that number at four for a cent; by selling them 
at twenty-five for 12 cents, I gained 18 cents, How many pears did 
I buy ? 


19, Solve the equations: 
a a _ 4%42 74+14 
(1) — alee 


e 


=5—6%+ 








9) X+Y 8z—5Y _o ada bate 
a ey az : cAlce 

90. Divide a2xa§+ (2ac—b?)x#+c? by axt+c—b2?. 

91, Ifa horses are worth b cows, and c cows are worth d sheep, 
find the value of a horse when a sheep is worth $2. 

99, Find the highest common factor of 8a7b3c, 12a%bc?, 15a2b5 ; 
and the lowest common multiple of 4ab2c3, 12a%b, 18ac?. 
aah oy IPOS MUS 

2ab2c 362 b2d =~ 6. a2? 

93, A gentleman divided $49 amongst 150 children. Each girl 
had 50 cents, and each boy 25 cents. How many boys were there ? 


94, Ii V=5a+4b—-6c, X=—8a—9b+7c, Y=20a+7b—85c, 
Z=18a—5b+4+9c, calculate the value of V—(X+ Y)+Z. 


Also find the value o 


MISCELLANEOUS EXAMPLES V, 179 


95, Solve the equations : 
l 3(6 — 5x) 63a _ 3a _ 3G 
() a) a KE Oy Rae 





2) plety)=e+1, Hy-n)=20-1. 


96, Find the factors of 
(1) a?-a-182; (2) 8a?+13x-6. 
Oi 4 eye AN ey, ye the value of 
Ni{5(y? — 22) — ah + V/Bfa(a? — 2?) — 1h. 
98, The product of two expressions is («+ 2y)?+ (3%+z)°, and one 
of them is 4%+2y+2; find the other. 


99, When 4 and B sit down to play, B has two-thirds as much 
money as A ; after a time A wins $15, and then he has twice as 
much money as 6. How much had each at first? 


80, Find the square root of 16a°+4a+4 -16a?+a?-8at. 
| 





81, Find the value of 
| 1 ae Teed pte 
2a {2 p(e3)} {a —3le42)} (e-F(2-5)}, 


and subtract the result from (a+ 2)(%—38)(~+4). 


382, Find the square root of 





a4 293 lla? 9 
Gs. PAG eG 


BO ela ciieiaD =.c= es = 1, find the value of 
arc _ nt Jatd — Vaid + o 
me 


a 
34, Separate into he simplest factors : 
(1) 2?—xy—-6y’; (2) 2° —4xy?- ay +4y%. 


85, Solve the equations : 
(1) (a-—1)(a%-2)(%-6) = (w- 3); 
~ ome: 
(2) 2 5 3 


2a! By =0, ms ys 13. 
36, A farmer ere to one person 9 horses and 7 cows for $375, 
and to another 6 horses and 13 cows at the same prices and for the 
same sum: what was the price of each ? 


180 ALGEBRA. 


37, When a=3, b=2, c=—7, find the value of 
(1) 3Bb%c | Bbc? _ Ga? ; 
a a-c b+2c¢ 
(2) 4c+{c-(3¢-—2b) + 2b}. 
88, Solve the equations : 
1,1 1 





(2) 542+3y = 120, 10.c = Sy + 90. 


39, Find the highest common factor of 5a°+2z?-—15a2-—6 and 
7x? — 447 — 21a +12. 


40, A coach travels between two places in 5 hours; if its speed 
were increased by 3 miles an hour, it would take 34 hours for the 
journey : what is the distance between the places ? 


41, From 2(a+a-b)(c-—a+b) take (%-a)(x%—-b)(vt+a+tDd). 
42, Find the value of 
2a*+5a~3 3 3a? —102+3, 62? -—52+1 
a3 — 9x 2+ 3042 © Bae +7a%+2a 








43, Divide 2°+y?+3xy-1 by x+y-—-1, and extract the square 
root of «4 - 3.° +5 4 204% 


44, Aman can walk from A to B and back in a certain time at 
the rate of 4 miles an hour. If he walks at the rate of 3 miles an 
hour from A to B, and at the rate of 5 miles an hour from B to A, 
he requires 10 minutes longer for the double journey. What is the 
distance from A to B? 

45, Find the highest common factor of 

7x4 — l0ax® + 8a72? - 4da®x+4at, 8rt— 13az3+ 5a2x? - 3a%a + 3a'. 


46. Solve the equations : 


(2) w-2y4+z2=0,° 9%-8y+32=0, 22+3y+5z2 = 36. 
47, Find the lowest common multiple of 
6a?-a2-1, 3a7+7x4+2, 227+32—-2. 


MISCELLANEOUS EXAMPLES V., 181 


48, The expression ax+3b is equal to 30 when 2 is 3, and to 42 
when 2 is 7: what is its value when 2 is 1; and for what value of x 
is it equal to zero? 


49, Find the lowest common multiple of 
4(a?+ab), 12(ab?-b*), 18(a?—b?). 
50, Extract the square root of 


4a? _ 128, on | 24a , 16a? 
da 8G: x Figs 





51, Reduce to lowest terms 
1224 + 423 — 23x? — 9a — 9 
8x4 — 1427-9 
52. Solve the equations : 





OU CY ett OY Qu —Ty 3a —7 
1 a et se Fh a ee PP ames | Vireo : 
Gy 2— Fer em eg y 6 ial ae 
(2) 2e-y+32=1, 4¢+3y-2z2=13, 6x-4y+2=20. 
53. Simplify 
(1) GHD 2a 8 obo 
b a+b ab—b?’ 
ee 1 
24+ 8a+15 x?+1le+30 
54, The sum of the two digits of a number is 9 ; if the digits are 


reversed the new number is four-sevenths of what it was before. 
Find the number. 





(2) 








55, Solve the equations : 
(1) 4ar- (5y-4)=1, es 1-1, 
(2) Sx+dy-11l=0, 5y—-6z=-—8, 7z2-8x%-—13=0. 
56, Find the value of : 
2 ] 3x a 


atx a-x x -a@ (a+a)* 
57, Resolve into factors : 


A) 28 -2a* a; (2) af+at—ai—l1. 








182 ALGEBRA. 


58, Two persons started at the same time to go from A to B. 
One rode at the rate of 74 miles per hour and arrived half an hour 
later than the other who travelled by train at the rate of 30 miles 
per hour. What is the distance between A and B? 


59, Find the square root of 
Ae Loge ays 4 Gay 16a? 
9y2 2 l5yz 1622 52? 2527" 





60. Find the factor of highest dimensions which will exactly 
divide each of the expressions 


2c4+ c8d — c2d? --7cd?—4d4, 3c4+ c8d - 2c?d* — 9cd? — 5d4. 





: : 2 3 — Qn? 
Simplif Tye et ee ee 
G1, Simplify (1) —;| Fail Mee 


DACiie aler) 


62, Find the highest common factor of 
624 — 2a3+927+9"—4 and 924+ 80x? — 9. 


What value of x will make both these expressions vanish ? 








63. Solve the equations : 








9 293 
(1) x-3 x-6 
2¢ Se 1 ba 15> 

yi ~ SAN 
(2) rE NE eS x=2 


e 
9 





6x? — Say —6y? — 15x? + 8ay — 121? 
4x? — Bay +3y? 35a?+47xy + by” 





64, Simplify i 


65, Find the value of 


x-2a_x2+2a  Il6ab Shane 
e+2b 2-2b 462-2? .. 


4ab 
a+b 

66, An egg-dealer bought a certain number of eggs at 16 cents 
per score, and five times the number at 75 cents per hundred; he 


sold the whole at 10 cents per dozen, gaining $3.24 by the transac- 
tion. How many eggs did he buy ? 


MISCELLANEOUS EXAMPLES V. 183 
67, If a=-1, b=-2, c=-3, d=—4, find the value of 


2a? +ab?—8abe_a-b+c-d_a ” Ze 
a? — b?-—abc be — 2ad brad 








68, Solve the simultaneous equations : 
Dee TY ee al eZ 
<=) > 








2 14 8 4 
ey YO 5(y +1) 
MAA ite aE 
sau 710 Pe ay 
69, Simplify the fractions: 
(iy cae ey) 
x-%Z x-Yy (%-x)(y- 2) 
wig ut el 
5 B a b a 
eee) ee 
b a/\b a a2 hb? ab 


70, From a certain sum of money one-third part was taken and 
From the sum thus increased one-fourth part 


$50 put in its stead. 
If the amount was now $120, 


was taken and $70 put in its stead. 
find the original sum. 
71, Find the lowest common multiple of 
(a4 —a%c?)?, 4a®-Sate?+4a2ct, a+ 3a3c + 3a%c?+ ac’. 
72, Solve the equations: 
; ‘135% —°225_ 36 -09%-°18 , 
(1) *l5a+ 03 es, z= 9 ; 
ll - 5a, da —-33 
6 








l0x+4, 7-20? _ 


(2) SAY EVER S as 





73, Simplify 
x?—44¢—-21 y C+ Ga? -247a , x? - 202491 
ae era 





e+17 x? —a2-12 


74, What must be the value of x in order that 
(a+2a)? 
a? + 70ax + 3.x? 


may be equal to 14 when a is equal to 67 ? 


184 ALGEBRA. 


75, Find the highest common factor of 1674+36a?+81 and 
8x? +27; and find the lowest common multiple of 82°+27, 
16a*+ 3627+81, and 62?-5x-6. 

76, Solve the equations : 

(1) a(a—a)-—b(a-—b)=(a+b)(x-a-—b). 
(2) (a+b)z--ay = a?, (a? + b*)x -— aby = a’. 
77. A farmer bought a certain number of sheep for $30. He 


sold all but five of them for $27, and made a profit of 20 per cent. 
on those he sold: find how many he bought. 


78, Find the value of 
ey vty 2a-8b 3b 


(1) oe eA ee eth gh (2) 2Qa—6b 2a 
a?—y? atty?? ~" 2a=3b 3b 


ory? 72 — 2a 2a —6b 

















79, Find the H.C.F. of 212? - 26a?+ 8a and 6a723 — a7a? — 2a°x. 
Also the L.C.M. of z?-2, ax?+2a2—-3a, x?-—72?+6z2. 
80, Simplify the expression 
(a+b+c)(a-b+c)-{(atcP—-b— (a+b +c")}. 
81, Solve the equations : 








T+u QWw-y 5Sy-7 , 4a-3 < 
ee nD : sis oF; 
(1) 5 p 3y — 9, 5 “b 6 Ov 
(2) Se hie anh) 

3u—-2Z x2+1 2243 
82, Extract the square root of 

aes 3003 ey: 

#4 V2 +4) 4+9(2 +4) ~4(Z 42) 45. 

Ope ee pe 3 ae JP 








83, Find the value of 
4a+6b | 6a-4b _4a°+65? , 4b?- 6a? , 200! 
a+b a—b a? — b? a+b? at—bt 








84, A bag contains 180 gold and silver coins of the value alto- 
gether of $144. Each gold coin is worth as many cents as there are 
silver coins, and each silver coin as many cents as there are gold 
coins. low many coins are there of each kind ? 


MISCELLANEOUS EXAMPLES V. 185 


85. Solve the equations: 
ibe ob Soaiom ke 
glo a+4 247? 
(2) V22+6-Nx—1=2. 
86, Find the factors of 
(1) 20a?+2lab-27b?; (2) a3-3z?--9x+27. 


87, If the length of a field were diminished and its breadth 
increased by 12 yards, it would be square. If its length were 
increased and its breadth diminished by 12 yards, its area would be 
15049 square yards. Find the area of the field. 

68, Simplify the expression 

{(a+b)(at+b+c)+c7{(a+b)? - 7} 
{(a+bP-C}{at+b+ey 
89, Find the square root of 
(2x + 1)(2a + 3)(2% + 5)(2%+7) + 16. 


90, Simplify 





10a? paket fe 2 
(1+a?)(l-4a2) 1-2a 1+a?™ 








91, Solve the equations: 





liye PER SNE ae 
ine ain me oe 
Oh ye mae ee 


99, Resolve into factors: 
(1) 622+52-6. (2) 9at-82ax°y?+9y4. 


a4 — 154? 4+- 28a - 12 


Red 
ea Ema Be 


to its lowest terms. 


94, Simplify the fraction 


(a+b)? a+ 2b+a , (a+ b)z ak 
(a-a)(xtat+b) Wwe-a) w#+be-a?-ah 2 


186 ALGEBRA. 


95, A person being asked his age replied, ‘‘Ten years ago I was 
five times as old as my son, but twenty years hence he will be half 
my age.” What is his age? 

96, Find the value of 

{ 2a, Lee ee eee 
(a-a)(ate) a-a wtasy az 








Oy. When ia yoo ee g d=-1, find the value of 


a’ — 03 —(a—b)> - 11(80 +20)(2e-F). 


a“ 


98, Find the square root of 


a* + b4 — a®b — ab? + 9a" 


99, Solve the equations: 


(1) 3x 24 


= =92- 9 2 = 493. 
a OeeT 4 (2) 327+ 227 = 493 











LOO DEY Negi onan ined 
101, What value of x will make the sum of pear and Hane 
equal to 2? 


102, Aman drives toacertain place at the rate of 8 miles an hour ; 
returning by a road 3 miles longer at the rate of 9 miles an hour he 
takes 74 minutes longer than in going: how long is each road ? 


103, Find the product of 
(1) 32°?-4ay+ Ty’, 3a*+4ay+ Ty? ; 
(2) #?-2y7, 2?-Qay+2Qy?, x?+2y?, 2? +2ry+2y?. 
104, Extract the square root of 
309 


1-50? 4-208 + Sat - S054 a 


MISCELLANEOUS EXAMPLES V. 187 


105, Find the highest factor common to 
x(62? — 8y?) — y(3a?-4y?) and Qay(2y —x) +423 — 2y3, 


106, The sum of the digits of a number is 9, and if five times the 
digit in the tens’ place be added to twice the digit in the units’ 
place, the number will be inverted. What is the number? 


2 
the yeeaaes (a+1) = 3, prove that a® neo. 
(eR? 


108, Solve the equations : 
(1) (x+7)(y-3)+2-(y+8)(x-1) =5x-1ly+35=0; 


(2) pastes) = 
3 64-32 3(x-2) 








109, If x=b+c, y=c-a, z=a-b), find the value of 
x+y? +27 — Qay — Quea+ Qyz. 
110, Express in the simplest form 


en Leer Pee tee . a 
02+ 3e+1 6a2+bet+1] 1222+7x24+1 2022+9xe41 





11]. Resolve into factors : 
(1) 4a7b? — (a2 +b? -c?)*; 
(2) ab(m?+1)+m(a? +b”). 


112, Simplify the fractions : 





] a-x 
——— at Nee 
(1) —: @) — 
ete CH= the 
14.21 ie 
3-2 x 


1138, Solve the equations : 
75-2 , 802+21_ 23 
1 : = 55 
(1) 3(e41) 5(8a42) £41 


(2) Ve+12— Ja =6. 





188 ALGEBRA. 


114, Ten minutes after the departure of an express train a slow 
train is started, travelling on an average 20 miles less per hour, 
which reaches a station 250 miles distant 34 hours after the arrival 
of the express. Find the rate at which each train travels. 


115, Simplify 
a’ — 64a , [ 2a27+5a4+2. ee 
a*—4 °2a?+9a+4° \a?+4a° a?+a—2 ; 








116. Resolve into four factors 
4(ab + cd)? — (a? +b? - c? - d’)?. 


LL When Oa 4 OH ec :, d=-1, find the numerical 


value 





4c? — ala — 2b — d) —‘Vb4e + 11b3a?. 
118, Find the value of 





119, Solve the equations : 


120, A has 19 miles to walk. At the end of a quarter of an 
hour he is overtaken by B who walks half a mile per hour faster ; 
by walking at the same rate as & for the remainder of the journey 
he arrives half an hour sooner than he expected. Find how long the 
journey occupied each man. 


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